﻿ Exact Trigonometric Values from 1° to 90°

# Exact Trigonometric Values from 1° to 90°

Trigonometric functions are based on a right-angled triangle. But most of angles from 1° to 90° are not obvious at a glance of their trigonometric values except some special angles. It looks difficult and unnecessary to derive the exact values of all these values considering the convenience of using a calculator and trigonometric table chart. It's fun to navigate across multiple mathematical fields and find the mysterious numbers.

There are six functions of an angle commonly used in trigonometry, sine (sin), cosine (cos), tangent (tan), cotangent (cot), secant (sec), and cosecant (csc). If one of them has been determined, the exact values of other trig functions could be computed using trigonometric identities. For some special angles, such as 15°, 18°, 30°, 36°, 45°, 72°, their trigonometric values could be determined by using some special triangles.

## 30°

Trigonometrical values of 30° angle could be determined in a 30-60-90 triangle, in which one of its non-right internal angle is 30° and the other one is 60° in measure. The special triangle could be made by cutting an equilateral triangle in half along the altitude of any side. Then it's obvious the opposite side of 30° angle is the half of the hypotenuse in length. Hence,

\sin 30° = \dfrac{1}{2} and \cos 30° = \dfrac{\sqrt{3} }{2}

In the meantime, the trigonometrical values of 60° angle also become known values.

\sin 60° = \dfrac{\sqrt{3} }{2} and \cos 60° = \dfrac{1}{2}

## 15°

Since 15 is the half of 30, \sin 30° could be determined by using the following half angle identity. Negative case is cancelled as sines values of all angles in quadrant I are positive.

\sin\dfrac{ \alpha}{2} = \sqrt{\dfrac{1-\cos \alpha }{2} }

The trig values of 15° angle can also be determined geometrically.

Extend BC to D such that CD = AC.

\sin 15° = \dfrac{AB}{AD} =\dfrac{AB }{\sqrt{AB^2+BD^2} }

=\dfrac{1}{2}\cdotp\sqrt{2-\sqrt{3} } =\dfrac{\sqrt{6}-\sqrt{2} }{4}

\cos 15° = \dfrac{BC}{AD} =\dfrac{1}{2}\cdotp\sqrt{2+\sqrt{3} } =\dfrac{\sqrt{6}+\sqrt{2} }{4}

Using the co-function identities, sines and coses values of 75° angle are also determined.

\sin 75°=\sin (90°-15°)=\cos 15°

\cos 75°=\cos (90°-15°)=\sin 15°

## 45°

Trigonometric values of 45° angle could be easily obtained in a 45-45-90 triangle. Since the two non-right internal angles are equal to each other, it's an isosceles right triangle. According to Pythagorean Theorem, the hypotenuse is \sqrt{2} in length if the length of the opposite and adjacent sides is 1. Then, we get,

\sin 45° = \cos 45°=\dfrac{1}{\sqrt{2} }= \dfrac{\sqrt{2} }{2}

## 18°

The angle of 18° is just the half angle of the vertex angle of a 36-72-72 triangle, which is also called golden triangle in that the ratio of its long side to base side is equal to golden radio.

Drop an altitude to the base, and the sines and cosines values of 18° are computed as

\sin18\degree =\dfrac{-1+\sqrt{5} }{4} , \cos 18\degree =\dfrac{\sqrt{10+2\sqrt{5} } }{4}

Using the co-function identities, the sines and coses values of 72° angle are also determined. Using double angle identity, it's easy to compute the trig values of 36° angle.

\cos 36°= \dfrac{1+\sqrt{5} }{4}, \sin 36° =\dfrac{1}{4}\cdotp \sqrt{10-2\sqrt{5} }

In the meantime, the trig values for 54° are revealed using co-function identities.

## 3°

Up to now, we have been able to work out the trigonometric values of the following angles by using some of special triangles. One of the common feature of these angles is that they all are divisible by 3, which reminds us 3° is probably our next target to find its trigonometric values.

\begin{array}{|c|c|c|c|c|c|c|c|c|c} \hline 15°&18°&30°&36°& 45°&54°&60°&72°& 75° \\ \hline \end{array}

Noticed 3 is the difference of 18 and 15. By using difference identity, the values of sin 3° and cos 3° could be determined.

\sin 3\degree = \sin(18\degree-15\degree)

=\sin 18\degree\cos 15\degree-\cos18\degree\sin 15\degree =\dfrac{(\sqrt{5}-1)(\sqrt{6}+\sqrt{2} )-\sqrt{10+2\sqrt{5} }(\sqrt{6}-\sqrt{2})}{16}

\cos 3° = \cos(18° -15° )

= \cos 18°\cos 15° +\sin 18° \sin 15°

=\dfrac{(\sqrt{6}+\sqrt{2} )\sqrt{10+2\sqrt{5} }}{4}+\dfrac{(\sqrt{5}-1)(\sqrt{6}-\sqrt{2})}{4}

With the complement of trigonometric values of 3°, all angles that are multiples of 3 within 90 degrees can be computed by using sum identities or half identities.

\sin(\alpha +\beta ) = \sin \alpha\cos \beta +\cos \alpha \sin \beta

\sin(\alpha -\beta ) = \sin \alpha\cos \beta -\cos \alpha \sin \beta

\cos(\alpha +\beta ) = \cos \alpha\cos \beta -\sin \alpha \sin \beta

\cos(\alpha -\beta ) = \cos \alpha\cos \beta +\sin \alpha \sin \beta

\begin{array}{|c|c|c|} \hline 6°&36°-30°& 2\times 3° \\ \hline 9°&45°-36°&36°-30° \\ \hline 12°&72°-60°& \\ \hline 21°&18°+3°&75°-54° \\ \hline 24°&54°-30°& \\ \hline 27°&72°-45°& \\ \hline 33°&18°+15°& \\ \hline 39°&54°-15°& \\ \hline 42°&60°-18°& \\ \hline 48°&30°+18°& \\ \hline 51°&36°+15°& \\ \hline 57°&60°-3°& \\ \hline 63°&60°+3°& \\ \hline 66°&36°+30°& \\ \hline 69°&72°-3°& \\ \hline 78°&60°+18°& \\ \hline 81°&45°+36°& \\ \hline 84°&54°+30°& \\ \hline 87°&90°-3°& \\ \hline \end{array}

Now there leave those that are not multiples of 3 to be determined. Especially, the exact values of sin 1° and cos 1°, which may be of interest for many people. But the involves triple identities and solving cubic equations.

## Derivation of triple identities

Triple identity for both sines and cosines functions are obtained by using the sum, double and Pythagorean identities.

\sin 3\alpha = \sin (\alpha +2\alpha ) \\ =\sin \alpha \cos 2\alpha +\cos \alpha \sin 2\alpha \\ =\sin \alpha (1-2\sin^2\alpha )+2\sin \alpha \cos^2\alpha \\ =\sin \alpha - 2\sin^3\alpha +2\sin \alpha (1-\sin^2\alpha ) \\ =3\sin \alpha-4\sin^3\alpha

\cos 3\alpha =\cos(\alpha +2\alpha ) \\ =\cos \alpha \cos 2\alpha -\sin \alpha \sin 2 \alpha \\ =\cos \alpha (2\cos^2 \alpha -1)-2\sin^2\alpha \cos \alpha\\ =2\cos^3 \alpha -\cos \alpha -2(1-\cos^2\alpha )\cos \alpha \\ =2\cos^3 \alpha -\cos \alpha -2\cos \alpha +2\cos^3 \alpha \\ =4\cos^3 \alpha -3\cos \alpha

Now we get the triple identities for both sines and cosines functions

\sin 3\alpha =3\sin \alpha-4\sin^3\alpha
(1)
\cos 3\alpha =4\cos^3 \alpha -3\cos \alpha
(2)

It appears to obtain the value of sines or cosines of 1° is to solve a cubic equation. Not only for 1°, but other angles including those are multiples of 3 whose trigonometric values could be determined by this method.

## Solving the depressed cubic equation

The cubic equation without the quadratic term is in the form that is called depressed cubic equation.

x^3+px+q=0
(3)

This equation was first released by Italian mathematician Girolamo Cardano. But Cardano's method gave only deal with the solutions with direct real numbers because he was not aware of complex number in that era. By introducing complex number, any cubic equation has three solutions. For example, the simplest cubic equation x^3=1.

Move 1 to the left side,

x^3-1=0

Apply the difference of cubic formula,

(x-1)(x^2+x+1)=0

Then, we get one real solution and one quadratic equation

x- 1=0 \to x=1 or x^2+x+1=0

x=\dfrac{-1\pm\sqrt{1-4} }{2} =\dfrac{-1\pm i\sqrt{3} }{2}

We get 3 cube roots of 1 for the simple cubic equation, one real number and two conjugate imaginary numbers, namely, 1, \dfrac{-1+i\sqrt{3} }{2} and \dfrac{-1-i\sqrt{3}}{2}

Then how do we solve generic form of the depressed cubic equation? We are going to solve the equation by using factorization method.

Let

x=u-\dfrac{p}{3u}
(4)

Substituting (4) into (3) transforms the depressed cubic equation to a quadratic equation in terms of u^3. Once the intermediate variable u is determined, x will be the sum of u and the product of -\dfrac{p}{3} and the reciprocal of u.

\Big( u-\dfrac{p}{3u}\Big) ^3+p\Big( u-\dfrac{p}{3u}\Big) +q=0

u^3-up+\dfrac{p^2}{3u}-\dfrac{p^3}{27u^3}+up-\dfrac{p^2}{3u}+q=0 Rearranging the terms, we get an identical depressed cubic equation like (3)

x^3+3utx+t^3-u^3=0
(5)

Solving the equation shows that

u^3=\dfrac{-q\pm\sqrt{q^2+\dfrac{4p^3}{27} } }{2} =-\dfrac{q}{2}\pm\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }
(6)

Taking cube root will give the value of u.

Considering the root with positive sign before the radical,

u = \sqrt{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } ,

the second term of (4) is

-\dfrac{p}{3u} = -\dfrac{p}{3}\cdotp \dfrac{1}{\sqrt{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } } =\sqrt{-\dfrac{q}{2}-\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } }

which shows that radicants of both terms of x are conjugate if the discriminant is less than 0.

Introducing complex number, 3 solutions are found for the depressed cubic equation after evaluating 6 cases.

x =\begin{cases} \sqrt{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } +\sqrt{-\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }}& \\ \\ ω\cdotp \sqrt{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } + \overline{ω} \sqrt{-\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }}&\ \\ \\ \overline{ω}\cdotp \sqrt{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } + ω\cdotp \sqrt{-\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }} \end{cases}
(7)

in which,

ω = \dfrac{-1+i\sqrt{3}}{2} and \overline{ω} =\dfrac{-1-i\sqrt{3}}{2}

It is noted that the radicant under the cube root is the discriminant to determine roots distribution of the equation.

\Delta = \dfrac{q^2}{4}+\dfrac{p^3}{27}

If \Delta < 0 , the equation has three real roots even though the redicants under the cube root radicals of both the first and the second terms are imaginary numbers. But they are conjugate and will remain conjugate after taking cube root. In the mean time, ω and \overline{ω} are also conjugate complex numbers. Let's find what impact after multiplication of the two pairs of conjugate complex numbers.

Let z_1 = a+bi, then \overline{z_1} = a-bi

Let z_2 =c+di, then \overline{z_2} = c-di

The first case

z_1\overline{z_2}+\overline{z_1}z_2

=( a+bi)(c-di)+(a-bi)(c+di)

=2ac+2bd

The second case

z_1z_2+\overline{z_1}\overline{z_2}

=( a+bi)(c+di)+(a-bi)(c-di)

=2ac-2bd

It's found that the sums of multiplications of two pairs of complex numbers in both cases are real numbers. Of course, the sum of two conjugate complex numbers is also real number. That's why the solutions are real numbers if the discriminant of a cubic equation is less than 0 even though there appear complex numbers in the formula of solutions. This conclusion will be used to determine the roots for \sin 1° and \sin 10°

## 1°

Since 1° is one third of 3°, given the value of \sin 3°, a cubic equation in terms \sin 1° is established by using the triple identity.

\sin^3 1°-\dfrac{3}{4} \sin 1°+\dfrac{\sin 3°}{4} =0

The equation is in perfect form of depressed cubic equation as (3), where

x = \sin 1°

p=-\dfrac{3}{4}

q=\dfrac{\sin 3°}{4}

\Delta =\dfrac{q^2}{4}+\dfrac{p^3}{27} = \dfrac{1}{4}\cdotp \dfrac{\sin^2 3\degree }{16} -\dfrac{1}{27}\cdotp \Big( \dfrac{3}{4} \Big) ^3=\dfrac{1}{64}\Big( \sin^2 3\degree -1 \Big) =-\dfrac{\cos^2 3\degree }{64} <0

Using the cube root formula directly gives

x_0=\dfrac{ \sqrt{-\sin 3°+i\cos 3° } +\sqrt{-\sin 3°-i\cos 3°}}{2}

x_1=\dfrac{ω \sqrt{-\sin 3°+i\cos 3° } +\overline{ω}\sqrt{-\sin 3°-i\cos 3°}}{2}

x_2=\dfrac{ \overline{ω} \sqrt{-\sin 3°+i\cos 3° } +ω\sqrt{-\sin 3°-i\cos 3°}}{2}

But the problem is that we have three solutions for \sin 1° here, which seems not right for an angle in the first quadrant. What's wrong with our computation? Well, we will need to start with multiplication of complex numbers.

A complex number can be represented by a point in the complex plane. It can be expressed in polar representation or modulus and argument form.

z = r \cos θ + ir \sin θ

in which r is called modulus of the complex number. It is the length of the line segment from the point to origin. θ is called argument of the complex number. It represents the angle from the positive axis to the line segment.

Suppose w is another complex number, which has modulus of s and argument φ

w = s \cos φ + is \sin φ

Multiplication of z and w gives

zw = (r \cos θ + ir \sin θ)( s \cos φ + is \sin φ)

=rs\cos θ \cos φ + i rs \cos θ \sin φ+i rs\sin θ\cos φ-rs\sin θ\sin φ

=(rs\cos θ \cos φ-rs\sin θ\sin φ)+i (rs \cos θ \sin φ+rs\sin θ\cos φ)

=rs\cos(θ+φ)+i rs\sin(θ+φ)

which says the product of two complex numbers yields a new complex number with its modulus equal to the product of the modulus of two complex numbers and its argument equal to the sum of the argument of two complex numbers.

If z=w, we get the formula for square of a complex number.

z^2 = r^2\cos 2θ+i r^2\sin 2θ

Multiplying by the complex number one more time gives the formula for cube of a complex number.

z^3 = r^3\cos 3θ+i r^3\sin 3θ

Inverse operation gives three cube root solutions, which are spaced evenly around a circle.

z^{\frac{1}{3}}=\sqrt{r} \cos\dfrac{\theta +2k\pi }{3}+i\sqrt{r} \sin\dfrac{\theta +2k\pi }{3} , k=0, 1,2

x=\dfrac{ \sqrt{-\sin 3°+i\cos 3° } +\sqrt{-\sin 3°-i\cos 3°}}{2}

=\dfrac{ \sqrt{-\cos 87°+i\sin 87° } +\sqrt{-\sin 87°-i\sin 87°}}{2}

=\dfrac{ \sqrt{\cos (180°-87°)+i\sin(180°-87°) } +\sqrt{ \cos (180°-87°)-i\sin (180°-87°)}}{2}

=\dfrac{ \sqrt{\cos 93°+i\sin 93° } +\sqrt{ \cos 93°-i\sin 93°}}{2}

=\dfrac{ \sqrt{\cos 93°+i\sin 93° } +\sqrt{ \cos (-93°)+i\sin(- 93°)}}{2}

=\dfrac{\cos\dfrac{93°+360°k}{3}+i\sin \dfrac{93°+360°k}{3} + \cos\dfrac{-93°+360°k}{3}+i\sin \dfrac{-93°+360°k}{3} }{2}

=\dfrac{\cos(31°+120°k)+i\sin(31°+120°k)+ \cos(-31°+120°k)+i\sin(-31°+120°k) }{2}

If k = 0,

x_0 = \dfrac{\cos31°+i\sin 31°+\cos(-31°)+i\sin(-31°)}{2} = \cos 31°

If k =1

x_1 = \dfrac{\cos151°+i\sin 151°+\cos 89° +i\sin 89° }{2} = \cos 31°\cos 120°+i2 \cos 31°\sin 120°

The result of x_2 looks inconsistent with our precious conclusion considering the discriminant is less than 0 here. What happens here is that we take cube roots of the first and second term independently from the expression of the final solution. If we rewrite the formula for the solution of the cubic equation from the equation (4),

x = \sqrt{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } - \dfrac{p}{3}\cdotp \dfrac{1}{ \sqrt{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } }}

Plug in p=-\dfrac{3}{4} and q=\dfrac{\sin 3°}{4} , then

x = \dfrac{ \sqrt{-\sin 3°+i\cos 3° } }{2}+\dfrac{1}{4}\cdotp \dfrac{1}{ \dfrac{ \sqrt{-\sin 3°+i\cos 3° } }{2}}

=\dfrac{1}{2}\Big( \sqrt{-\sin 3°+i\cos 3° }+\dfrac{1}{ \sqrt{-\sin 3°+i\cos 3° }} \Big)

=\dfrac{1}{2}\Big( \sqrt{\cos 93°+i\sin 93°}+\dfrac{1}{ \sqrt{\cos 93°+i\sin 93°}} \Big)

=\dfrac{1}{2}\Big( \cos(31°+ 120°k)+i\sin(31°+120k) +\dfrac{1}{\cos(31°+ 120°k)+i\sin(31°+120k) } \Big)

If k=0, x_0 = \cos 31°=\sin59°

If k = 1, x_1 = \cos 151° = -\sin 61°

If k=2, x_2 =\cos 271° =\sin 1°

It shows that the three solutions of the cubic equation represent sines values of 3 integer angles within 90° (For second solution, take its absolute value.), in which the third root is the value of \sin 1°.

Similarly, we can establish a cubic equation using the triple identity for cosines function.

x^3 -\dfrac{3}{4} x- \dfrac{1}{4} \cos 3° =0

Here,

p = -\dfrac{3}{4}

q = - \dfrac{1}{4} \cos 3°

Plug in the solution formula of a cubic equation,

x_0=\dfrac{1}{2}(\sqrt{\cos 3°+i\sin 3°}+\sqrt{\cos 3°-i\sin 3°})

x_1=\dfrac{1}{2}(ω\sqrt{\cos 3°+i\sin 3°}+\overline{ω}\sqrt{\cos 3°-i\sin 3°})

x_2=\dfrac{1}{2}(\overline{ω}\sqrt{\cos 3°+i\sin 3°}+ω\sqrt{\cos 3°-i\sin 3°})

Of course, there's only one solution representing the value of \cos 1°. Let's show that which angles they represent.

Rewriting the solution according to the equation (4) gives

x=\dfrac{1}{2}\Big( \sqrt{\cos 3°+i\sin 3°}+\dfrac{1}{ \sqrt{\cos 3°+i\sin 3°}} \Big)

=\dfrac{1}{2}\Big( \cos(1°+ 120°k)+i\sin(1°+120k) +\dfrac{1}{\cos(1°+ 120°k)+i\sin(1°+120k) } \Big)

If k=0, x_0 = \cos 1°

If k = 1, x_1 = \cos 121° = -\cos 59°

If k=2, x_2 =\cos 241° =-\cos 61°

It shows that the first root represent the value of \cos 1°. The other two are negative numbers of \cos 59° and \cos 61°

## 2°

How to find the exact values of \sin 2\degree or \cos 2\degree ? It looks obvious to compute it by using double angle identities since we have determined the value of \sin 1\degree. But this method involves multiplication of two complex numbers, especially under the cube root radicals. There are usually multiple methods to obtain the expressions, which may look differently but have the same value. Here we are looking for a way to reduce the complexity of the complex number operation. For instance,

2° = 72°-70°

When computing the trig values for 10°, we also obtain the exact values for \sin 70° and \cos 70° , which are expressed in the form of complex numbers. 72° is a special angle which can be determined geometrically by using golden triangle. Therefore, trig values of 2° could be computed by using the difference identities and given values.

\sin 2° = \sin(72° -70° ) = \sin 72°\cos 70° -\cos 72° \sin 70°

\cos 2°=\cos(72° -70° ) = \cos 72°\cos 70° +\sin 72° \sin 70°

\cos72\degree =\dfrac{-1+\sqrt{5} }{4} , \sin 72\degree =\dfrac{\sqrt{10+2\sqrt{5} } }{4}

\sin 70°=\dfrac{ ω\sqrt{\sin 30°-i\cos 30° } +\overline{ω}\sqrt{\sin 30°+i\cos 30°}}{2}

\cos 70°=\dfrac{ \overline{ω}\sqrt{ \sin 30°-i\cos 30° } +ω\sqrt{ \sin 30°+i\cos 30°}}{2}

The operation will only involve multiplication of a complex number and a constant. Substituting the trig values of 72° and 70° gives

\sin 2° =\dfrac{1}{8}\bigg[ \Big( \sqrt{10+2\sqrt{5} } \Big) \cdotp \Big( \overline{ω}\sqrt{\sin 30°-i\cos 30° } +ω\sqrt{\sin 30°+i\cos 30°}) -\Big( -1+\sqrt{5} \Big) \Big( ω\sqrt{\sin 30°-i\cos 30° } +\overline{ω}\sqrt{\sin 30°+i\cos 30°}\Big) \bigg]

\cos 2°=\dfrac{1}{8}\cdotp \bigg[ \Big( {-1+\sqrt{5}\Big) \Big( \overline{ω}\sqrt{ \sin 30°-i\cos 30° } +ω\sqrt{ \sin 30°+i\cos 30°}}\Big) -\sqrt{10+2\sqrt{5} } \Big( ω\sqrt{\sin 30°-i\cos 30° } +\overline{ω}\sqrt{\sin 30°+i\cos 30°}\Big) \bigg]

## 4°

Similarly, we are going to use the turnaround to reduce the complexity of complex number operation for 4°.

4° = 54°-50°

The exact values for \sin 4° and \cos 4° could be computed by using the difference identities and given values.

\sin 54°= \dfrac{1+\sqrt{5} }{4}, \cos 54° =\dfrac{1}{4}\cdotp \sqrt{10-2\sqrt{5} }

\sin 50\degree = \dfrac{ \sqrt{-\sin 30°+i\cos 30° } +\sqrt{-\sin 30°-i\cos 30°}}{2}

\cos 50\degree =- \dfrac{1}{2}(ω\sqrt{\cos 30°+i\sin 30°}+\overline{ω}\sqrt{\cos 30°-i\sin 30°})

\sin 4°=\sin(54° -50° ) = \sin 54°\cos 50° -\cos 54° \sin 50°

=\dfrac{1}{8}\bigg[ -\Big( 1+\sqrt{5}\Big) \Big( ω\sqrt{\cos 30°+i\sin 30°}+\overline{ω}\sqrt{\cos 30°-i\sin 30°}\Big) - \sqrt{10-2\sqrt{5} }\Big( \sqrt{-\sin 30°+i\cos 30° } +\sqrt{-\sin 30°-i\cos 30°}\Big) \bigg]

\cos 4°=\cos(54° -50° ) = \cos 54°\cos 50° +\sin 54° \sin 50°

=\dfrac{1}{8}\bigg[ -\sqrt{10-2\sqrt{5} }\Big( ω\sqrt{\cos 30°+i\sin 30°}+\overline{ω}\sqrt{\cos 30°-i\sin 30°}\Big) +(1+\sqrt{5} \Big) \Big( \sqrt{-\sin 30°+i\cos 30° } +\sqrt{-\sin 30°-i\cos 30°}\Big) \bigg]

## 5°

Using the triple identity for sines function, we get 3 solutions for the cubic equation.

x_0=\dfrac{ \sqrt{-\sin 15°+i\cos 15° } +\sqrt{-\sin 15°-i\cos 15°}}{2}

x_1=\dfrac{ω \sqrt{-\sin 15°+i\cos 15° } +\overline{ω}\sqrt{-\sin 15°-i\cos 15°}}{2}

x_2=\dfrac{ \overline{ω} \sqrt{-\sin 15°+i\cos 15° } +ω\sqrt{-\sin 15°-i\cos 15°}}{2}

To identify which angles they represent, we rewrite the solution according to equation (4). Then,

x = =\dfrac{1}{2}\Big( \sqrt{-\sin 15°+i\cos 15° }+\dfrac{1}{ \sqrt{-\sin 15°+i\cos 15° }} \Big)

=\dfrac{1}{2}\Big( \sqrt{\cos 105°+i\sin 105°}+\dfrac{1}{ \sqrt{\cos 105°+i\sin 105°}} \Big)

=\dfrac{1}{2}\Big( \cos(35°+ 120°k)+i\sin(35°+120k) +\dfrac{1}{\cos(35°+ 120°k)+i\sin(35°+120k) } \Big)

If k=0, x_0 = \cos 35°=\sin65°

If k = 1, x_1 = \cos 155° = -\sin 65°

If k=2, x_2 =\cos 275° =\sin 5°

It's found the third solution representing the value of \sin 5°, the first solution representing the value of \sin 65° and the second representing the negative value of \sin 65°

Similarly, establish a cubic equation for solving \cos 5° using the triple identity for cosines function which its cube roots given below,

x_0=\dfrac{1}{2}(\sqrt{\cos 15°+i\sin 15°}+\sqrt{\cos 15°-i\sin 15°})

x_1=\dfrac{1}{2}(ω\sqrt{\cos 15°+i\sin 15°}+\overline{ω}\sqrt{\cos 15°-i\sin 15°})

x_2=\dfrac{1}{2}(\overline{ω}\sqrt{\cos 15°+i\sin 15°}+ω\sqrt{\cos 15°-i\sin 15°})

Rewrite the solution according to the equation (4) to determine which angle they represent.

x=\dfrac{1}{2}\Big( \sqrt{\cos 15°+i\sin 15°}+\dfrac{1}{ \sqrt{\cos 15°+i\sin 15°}} \Big)

=\dfrac{1}{2}\Big( \cos(5°+ 120°k)+i\sin(5°+120k) +\dfrac{1}{\cos(5°+ 120°k)+i\sin(5°+120k) } \Big)

If k=0, x_0 = \cos 5°

If k = 1, x_1 = \cos 125° = -\cos 55°

If k=2, x_2 =\cos 245° =-\cos 65°

It shows that the first root represent the value of \cos 5°. The other two are negative numbers of \cos 55° and \cos 65°

## 6°

Using the triple identity for sines function, we get 3 solutions for the cubic equation.

x_0=\dfrac{ \sqrt{-\sin 18°+i\cos 18° } +\sqrt{-\sin 18°-i\cos 18°}}{2}

x_1=\dfrac{ ω \sqrt{-\sin 18°+i\cos 18° } +\overline{ω}\sqrt{-\sin 18°-i\cos 18°}}{2}

x_2=\dfrac{ \overline{ω} \sqrt{-\sin 18°+i\cos 18° } +ω\sqrt{-\sin 18°-i\cos 18°}}{2}

To identify which angles they represent, we rewrite the solution according to equation (4). Then,

x = =\dfrac{1}{2}\Big( \sqrt{-\sin 18°+i\cos 18° }+\dfrac{1}{ \sqrt{-\sin 18°+i\cos 18° }} \Big)

=\dfrac{1}{2}\Big( \sqrt{\cos 108°+i\sin 108°}+\dfrac{1}{ \sqrt{\cos 108°+i\sin 108°}} \Big)

=\dfrac{1}{2}\Big( \cos(36°+ 120°k)+i\sin(36°+120k) +\dfrac{1}{\cos(36°+ 120°k)+i\sin(36°+120k) } \Big)

If k=0, x_0 = \cos 36°=\sin54°

If k = 1, x_1 = \cos 156° = -\sin 66°

If k=2, x_2 =\cos 276° =\sin 6°

It's found the third solution representing the value of \sin 6°, the first solution representing the value of \sin 54° and the second representing the negative value of \sin 66°

Similarly, establish a cubic equation for solving \cos 6° using the triple identity for cosines function which its cube roots given below,

x_0=\dfrac{1}{2}(\sqrt{\cos 18°+i\sin 18°}+\sqrt{\cos 18°-i\sin 18°})

x_1=\dfrac{1}{2}(ω\sqrt{\cos 18°+i\sin 18°}+\overline{ω}\sqrt{\cos 18°-i\sin 18°})

x_2=\dfrac{1}{2}(\overline{ω}\sqrt{\cos 18°+i\sin 18°}+ω\sqrt{\cos 18°-i\sin 18°})

Rewrite the solution according to the equation (4) to determine which angle they represent.

x=\dfrac{1}{2}\Big( \sqrt{\cos 18°+i\sin 18°}+\dfrac{1}{ \sqrt{\cos 18°+i\sin 18°}} \Big)

=\dfrac{1}{2}\Big( \cos(6°+ 120°k)+i\sin(6°+120k) +\dfrac{1}{\cos(6°+ 120°k)+i\sin(6°+120k) } \Big)

If k=0, x_0 = \cos 6°

If k = 1, x_1 = \cos 126° = -\cos 54°

If k=2, x_2 =\cos 246° =-\cos 66°

It shows that the first root represent the value of \cos 6°. The other two are negative numbers of \cos 54° and \cos 66°

## 7°

To find the trig values of 27°, we need to do one more step, which is calculating the trig values of 27°.

First we have determined the trig values for 72° .

\sin 72° =\dfrac{\sqrt{10+2\sqrt{5} } }{4}

\cos 72° = \dfrac{-1+\sqrt{5} }{4}

Then, determine the trig values for 27° by using the difference identities.

\sin 27° =\sin(72° -45° ) = \sin 72°\cos 45° -\cos 72° \sin 45°

=\dfrac{1}{4}[\sqrt{5+\sqrt{5} } -\sqrt{3-\sqrt{5} } ]

\cos 27° = \cos(72° -45° ) = \cos 72°\cos 45° +\sin 72° \sin 45°

=\dfrac{1}{4}[\sqrt{5+\sqrt{5} }+\sqrt{3-\sqrt{5} } ]

We have also determined the exact values for \sin 20° and \cos 20°, which are expressed in the form of complex number.

\sin 20°=\dfrac{ \overline{ω} \sqrt{-\sin 60°+i\cos 60° } +ω\sqrt{-\sin 60°-i\cos 60°}}{2}

\cos 20° = \dfrac{1}{2}(\sqrt{\cos 60°+i\sin 60°}+\sqrt{\cos 60°-i\sin 60°})

And then compute the trig values of 20° by using difference identity since 7° is the difference between 27°and 20°.

\sin 7°=\sin(27° -20° ) = \sin 27°\cos 20° -\cos 27° \sin 20°

=\dfrac{1}{8}[(\sqrt{5+\sqrt{5} } -\sqrt{3-\sqrt{5} } )(\sqrt{\cos 60°+i\sin 60°}+\sqrt{\cos 60°-i\sin 60°})-(\sqrt{5+ \sqrt{5} }+\sqrt{3- \sqrt{5} } )( \overline{ω} \sqrt{-\sin 60°+i\cos 60° } +ω\sqrt{-\sin 60°-i\cos 60°})]

\cos 7 ° = \cos(27° -20° ) = \cos 27°\cos 20° +\sin 27° \sin 20°

=\dfrac{1}{8}[(\sqrt{5+\sqrt{5} }+\sqrt{3-\sqrt{5} } )(\sqrt{\cos 60°+i\sin 60°}+\sqrt{\cos 60°-i\sin 60°})-(\sqrt{5+\sqrt{5} }-\sqrt{3-\sqrt{5} } )(\overline{ω} \sqrt{-\sin 60°+i\cos 60° } +ω\sqrt{-\sin 60°-i\cos 60°})]

## 8°

First determine the trig values for 24°.

\sin 54°= \dfrac{1+\sqrt{5} }{4}, \cos 54° =\dfrac{1}{4}\cdotp \sqrt{10-2\sqrt{5} }

Given sines and cosines values for the special angles of 54° and 30°, the trig values for 24° could be computed by using difference identities.

\sin 24° =\sin(54°-30° ) = \sin 54°\cos 30°-\cos 54° \sin 30°

=\dfrac{1}{8}\Big[ (1+\sqrt{5}) \cdotp\sqrt{3} -\sqrt{10-2\sqrt{5} } \Big]

\cos 24° =\cos(54°-30° ) = \cos 54°\cos 30°+\sin 54° \sin 30°

=\dfrac{1}{8}\Big( \sqrt{10-2\sqrt{5} } \cdotp \sqrt{3} - (1+\sqrt{5} ) \Big)

Using the triple identity for sines function, we get 3 solutions for the cubic equation for \sin 16°.

x_0=\dfrac{ \sqrt{-\sin 24°+i\cos 24° } +\sqrt{-\sin 24°-i\cos 24°}}{2}

x_1=\dfrac{ ω\sqrt{-\sin 24°+i\cos 24° } +\overline{ω}\sqrt{-\sin 24°-i\cos24°}}{2}

x_2=\dfrac{ \overline{ω} \sqrt{-\sin 24°+i\cos 24° } +ω\sqrt{-\sin 24°-i\cos 24°}}{2}

To identify which angles they represent, we rewrite the solution according to equation (4). Then,

x = =\dfrac{1}{2}\Big( \sqrt{-\sin 24°+i\cos 24° }+\dfrac{1}{ \sqrt{-\sin 24°+i\cos 24° }} \Big)

=\dfrac{1}{2}\Big( \sqrt{\cos 114°+i\sin 114°}+\dfrac{1}{ \sqrt{\cos 114°+i\sin 114°}} \Big)

=\dfrac{1}{2}\Big( \cos(38°+ 120°k)+i\sin(38°+120k) +\dfrac{1}{\cos(38°+ 120°k)+i\sin(38°+120k) } \Big)

If k=0, x_0 = \cos 38°=\sin 52°

If k = 1, x_1 = \cos 158° = -\sin 68°

If k=2, x_2 =\cos 278° =\sin 8°

It's found the third solution representing the value of \sin 8°, the first solution representing the value of \sin 52° and the second representing the negative value of \sin 68°

Similarly, establish a cubic equation for solving \cos 8° using the triple identity for cosines function which its cube roots given below,

x_0=\dfrac{1}{2}(\sqrt{\cos 24°+i\sin 24°}+\sqrt{\cos 24°-i\sin 24°})

x_1=\dfrac{1}{2}(ω\sqrt{\cos 24°+i\sin 24°}+\overline{ω}\sqrt{\cos 24°-i\sin 24°})

x_2=\dfrac{1}{2}(\overline{ω}\sqrt{\cos 24°+i\sin 24°}+ω\sqrt{\cos 24°-i\sin 24°})

Rewrite the solution according to the equation (4) to determine which angle they represent.

x=\dfrac{1}{2}\Big( \sqrt{\cos 24°+i\sin 24°}+\dfrac{1}{ \sqrt{\cos 24°+i\sin 24°}} \Big)

=\dfrac{1}{2}\Big( \cos(8°+ 120°k)+i\sin(8°+120k) +\dfrac{1}{\cos(8°+ 120°k)+i\sin(8°+120k) } \Big)

If k=0, x_0 = \cos 8°

If k = 1, x_1 = \cos 128° = -\cos 52°

If k=2, x_2 =\cos 248° =-\cos 68°

It shows that the first root represent the value of \cos 8°. The other two are negative numbers of \cos 52° and \cos 68°

## 9°

\cos 36°= \dfrac{1+\sqrt{5} }{4}, \sin 36° =\dfrac{1}{4}\cdotp \sqrt{10-2\sqrt{5} }

Using the difference identities

\sin 9° = \sin(45° -36° ) = \sin 45°\cos 36° -\cos 45° \sin 36°

=\dfrac{1}{8}[ (\sqrt{2} +\sqrt{10}) - \sqrt{20-4\sqrt{5} }]

\cos 9° = \cos(45° -36° ) = \cos 45°\cos 36° +\sin 45° \sin 36°

=\dfrac{1}{8}[(\sqrt{2} +\sqrt{10}) + \sqrt{20-4\sqrt{5} } ]

## 10°

Using the triple identity for sines function, we get 3 solutions for the cubic equation.

x_0=\dfrac{ \sqrt{-\sin 30°+i\cos 30° } +\sqrt{-\sin 30°-i\cos 30°}}{2}

x_1=\dfrac{ ω \sqrt{-\sin 30°+i\cos 30° } +\overline{ω}\sqrt{-\sin 30°-i\cos 30°}}{2}

x_2=\dfrac{ \overline{ω} \sqrt{-\sin 30°+i\cos 30° } +ω\sqrt{-\sin 30°-i\cos 30°}}{2}

To identify which angles they represent, we rewrite the solution according to equation (4). Then,

x = =\dfrac{1}{2}\Big( \sqrt{-\sin 30°+i\cos 30° }+\dfrac{1}{ \sqrt{-\sin 30°+i\cos 30° }} \Big)

=\dfrac{1}{2}\Big( \sqrt{\cos 120°+i\sin 120°}+\dfrac{1}{ \sqrt{\cos 120°+i\sin 120°}} \Big)

=\dfrac{1}{2}\Big( \cos(40°+ 120°k)+i\sin(40°+120k) +\dfrac{1}{\cos(40°+ 120°k)+i\sin(40°+120k) } \Big)

If k=0, x_0 = \cos 40°=\sin50°

If k = 1, x_1 = \cos 160° = -\sin 70°

If k=2, x_2 =\cos 280° =\sin 10°

It's found the third solution representing the value of \sin 10°, the first solution representing the value of \sin 50° and the second representing the negative value of \sin 70°

Similarly, establish a cubic equation for solving \cos 10° using the triple identity for cosines function which its cube roots given below,

x_0=\dfrac{1}{2}(\sqrt{\cos 30°+i\sin 30°}+\sqrt{\cos 30°-i\sin 30°})

x_1=\dfrac{1}{2}(ω\sqrt{\cos 30°+i\sin 30°}+\overline{ω}\sqrt{\cos 30°-i\sin 30°})

x_2=\dfrac{1}{2}(\overline{ω}\sqrt{\cos 30°+i\sin 30°}+ω\sqrt{\cos 30°-i\sin 30°})

Rewrite the solution according to the equation (4) to determine which angle they represent.

x=\dfrac{1}{2}\Big( \sqrt{\cos 30°+i\sin 30°}+\dfrac{1}{ \sqrt{\cos 30°+i\sin 30°}} \Big)

=\dfrac{1}{2}\Big( \cos(10°+ 120°k)+i\sin(10°+120k) +\dfrac{1}{\cos(10°+ 120°k)+i\sin(10°+120k) } \Big)

If k=0, x_0 = \cos 10°

If k = 1, x_1 = \cos 130° = -\cos 50°

If k=2, x_2 =\cos 250° =-\cos 70°

It shows that the first root represent the value of \cos 10°. The other two are negative numbers of \cos 50° and \cos 70°

## 11°

First determine the trig values for 33°.

\sin 15° =\dfrac{1}{2}\cdotp\sqrt{2-\sqrt{3} } =\dfrac{\sqrt{6}-\sqrt{2} }{4}

\cos 15° =\dfrac{1}{2}\cdotp\sqrt{2+\sqrt{3} } =\dfrac{\sqrt{6}+\sqrt{2} }{4}

\sin18\degree =\dfrac{-1+\sqrt{5} }{4} = \dfrac{\sqrt{6-2\sqrt{5} } }{4} ,

\cos 18\degree =\dfrac{\sqrt{10+2\sqrt{5} } }{4}

Given sines and cosines values for the special angles of 15° and 18°, the trig values for 33° could be computed by using sum identities.

\sin 33° =\sin(18° +15° ) = \sin 18°\cos 15° +\cos 18° \sin 15°

=\dfrac{1}{8}\Big( \sqrt{6-2\sqrt{5} }\cdotp \sqrt{2+\sqrt{3} }+\sqrt{10+2\sqrt{5} }\cdotp \sqrt{2-\sqrt{3} }\Big)

\cos 33° =\cos(18° +15° ) = \cos 18°\cos 15° -\sin 18° \sin 15°

=\dfrac{1}{8}\Big( \sqrt{10+2\sqrt{5} }\cdotp \sqrt{2-\sqrt{3} }+ \sqrt{6-2\sqrt{5} }\cdotp\sqrt{2+\sqrt{3} }\Big)

Using the triple identity for sines function, we get 3 solutions for the cubic equation for \sin 33°.

x_0=\dfrac{ \sqrt{-\sin 33°+i\cos 33° } +\sqrt{-\sin 33°-i\cos 33°}}{2}

x_1=\dfrac{ ω\sqrt{-\sin 33°+i\cos 33° } +\overline{ω}\sqrt{-\sin 33°-i\cos33°}}{2}

x_2=\dfrac{ \overline{ω} \sqrt{-\sin 33°+i\cos 33° } +ω\sqrt{-\sin 33°-i\cos 33°}}{2}

To identify which angles they represent, we rewrite the solution according to equation (4). Then,

x = =\dfrac{1}{2}\Big( \sqrt{-\sin 33°+i\cos 33° }+\dfrac{1}{ \sqrt{-\sin 33°+i\cos 33° }} \Big)

=\dfrac{1}{2}\Big( \sqrt{\cos 123°+i\sin 123°}+\dfrac{1}{ \sqrt{\cos 123°+i\sin 123°}} \Big)

=\dfrac{1}{2}\Big( \cos(41°+ 120°k)+i\sin(41°+120k) +\dfrac{1}{\cos(41°+ 120°k)+i\sin(41°+120k) } \Big)

If k=0, x_0 = \cos 41°=\sin49°

If k = 1, x_1 = \cos 161° = -\sin 71°

If k=2, x_2 =\cos 281° =\sin 11°

It's found the third solution representing the value of \sin 11°, the first solution representing the value of \sin 49° and the second representing the negative value of \sin 71°

Similarly, establish a cubic equation for solving \cos 11° using the triple identity for cosines function which its cube roots given below,

x_0=\dfrac{1}{2}(\sqrt{\cos 33°+i\sin 33°}+\sqrt{\cos 33°-i\sin 33°})

x_1=\dfrac{1}{2}(ω\sqrt{\cos 33°+i\sin 33°}+\overline{ω}\sqrt{\cos 33°-i\sin 33°})

x_2=\dfrac{1}{2}(\overline{ω}\sqrt{\cos 33°+i\sin 33°}+ω\sqrt{\cos 33°-i\sin 33°})

Rewrite the solution according to the equation (4) to determine which angle they represent.

x=\dfrac{1}{2}\Big( \sqrt{\cos 33°+i\sin 33°}+\dfrac{1}{ \sqrt{\cos 33°+i\sin 33°}} \Big)

=\dfrac{1}{2}\Big( \cos(11°+ 120°k)+i\sin(11°+120k) +\dfrac{1}{\cos(11°+ 120°k)+i\sin(11°+120k) } \Big)

If k=0, x_0 = \cos 11°

If k = 1, x_1 = \cos 131° = -\cos 49°

If k=2, x_2 =\cos 251° =-\cos 71°

It shows that the first root represent the value of \cos 11°. The other two are negative numbers of \cos 49° and \cos 71°

## 12°

\cos 72\degree =\dfrac{-1+\sqrt{5} }{4} , \sin 72\degree =\dfrac{\sqrt{10+2\sqrt{5} } }{4}

Using the difference identities

\sin 12° = \sin(72° -60° ) = \sin 72°\cos 60° -\cos 72° \sin 60°

=\dfrac{1}{8}[ \sqrt{10+2\sqrt{5} } -\sqrt{3}( -1+\sqrt{5} ) ]

\cos 12° = \cos(72° -60° ) = \cos 72°\cos 60° +\sin 72° \sin 60°

=\dfrac{1}{8}[ (-1+\sqrt{5} )+\sqrt{3}\sqrt{10+2\sqrt{5} } ]

## 13°

First determine the trig values for 39°.

\sin 54°= \dfrac{1+\sqrt{5} }{4}, \cos 54° =\dfrac{1}{4}\cdotp \sqrt{10-2\sqrt{5} }

\sin 15° =\dfrac{1}{2}\cdotp\sqrt{2-\sqrt{3} } =\dfrac{\sqrt{6}-\sqrt{2} }{4}

\cos 15° =\dfrac{1}{2}\cdotp\sqrt{2+\sqrt{3} } =\dfrac{\sqrt{6}+\sqrt{2} }{4}

Given sines and cosines values for the special angles of 54° and 15°, the trig values for 39° could be computed by using difference identities.

\sin 39° =\sin(54°-15° ) = \sin 54°\cos 15°-\cos 54° \sin 15°

=\dfrac{1}{8}\Big[ (1+\sqrt{5}) \cdotp\sqrt{2+\sqrt{3} } -\sqrt{10-2\sqrt{5} } \cdotp\sqrt{2-\sqrt{3} } \Big]

\cos 39° =\cos(54°-15° ) = \cos 54°\cos 15°+\sin 54° \sin 15°

=\dfrac{1}{8}\Big[ \sqrt{10-2\sqrt{5} } \cdotp \sqrt{2+\sqrt{3} } - (1+\sqrt{5} ) \cdotp\sqrt{2-\sqrt{3} } \Big]

Using the triple identity for sines function, we get 3 solutions for the cubic equation for \sin 16°.

x_0=\dfrac{ \sqrt{-\sin 39°+i\cos 39° } +\sqrt{-\sin 39°-i\cos 39°}}{2}

x_1=\dfrac{ ω\sqrt{-\sin 39°+i\cos 39° } +\overline{ω}\sqrt{-\sin 39°-i\cos39°}}{2}

x_2=\dfrac{ \overline{ω} \sqrt{-\sin 39°+i\cos 39° } +ω\sqrt{-\sin 39°-i\cos 39°}}{2}

To identify which angles they represent, we rewrite the solution according to equation (4). Then,

x = =\dfrac{1}{2}\Big( \sqrt{-\sin 39°+i\cos 39° }+\dfrac{1}{ \sqrt{-\sin 39°+i\cos 39° }} \Big)

=\dfrac{1}{2}\Big( \sqrt{\cos 129°+i\sin 129°}+\dfrac{1}{ \sqrt{\cos 129°+i\sin 114°}} \Big)

=\dfrac{1}{2}\Big( \cos(43°+ 120°k)+i\sin(43°+120k) +\dfrac{1}{\cos(43°+ 120°k)+i\sin(43°+120k) } \Big)

If k=0, x_0 = \cos 43°=\sin 47°

If k = 1, x_1 = \cos 163° = -\sin 73°

If k=2, x_2 =\cos 283° =\sin 13°

It's found the third solution representing the value of \sin 13°, the first solution representing the value of \sin 47° and the second representing the negative value of \sin 73°

Similarly, establish a cubic equation for solving \cos 13° using the triple identity for cosines function which its cube roots given below,

x_0=\dfrac{1}{2}(\sqrt{\cos 39°+i\sin 39°}+\sqrt{\cos 39°-i\sin 39°})

x_1=\dfrac{1}{2}(ω\sqrt{\cos 39°+i\sin 39°}+\overline{ω}\sqrt{\cos 39°-i\sin 39°})

x_2=\dfrac{1}{2}(\overline{ω}\sqrt{\cos 39°+i\sin 39°}+ω\sqrt{\cos 39°-i\sin 39°})

Rewrite the solution according to the equation (4) to determine which angle they represent.

x=\dfrac{1}{2}\Big( \sqrt{\cos 39°+i\sin 39°}+\dfrac{1}{ \sqrt{\cos 39°+i\sin 39°}} \Big)

=\dfrac{1}{2}\Big( \cos(13°+ 120°k)+i\sin(13°+120k) +\dfrac{1}{\cos(13°+ 120°k)+i\sin(13°+120k) } \Big)

If k=0, x_0 = \cos 13°

If k = 1, x_1 = \cos 133° = -\cos 47°

If k=2, x_2 =\cos 253° =-\cos 73°

It shows that the first root represent the value of \cos 13°. The other two are negative numbers of \cos 47° and \cos 73°

## 14°

First determine the trig values for 42°.

\sin18\degree =\dfrac{-1+\sqrt{5} }{4} = \dfrac{\sqrt{6-2\sqrt{5} } }{4} ,

\cos 18\degree =\dfrac{\sqrt{10+2\sqrt{5} } }{4}

Given sines and cosines values for the special angles of 60° and 18°, the trig values for 42° could be computed by using difference identities.

\sin 42° =\sin(60° -18° ) = \sin 60°\cos 18° -\cos 60° \sin 18°

=\dfrac{1}{8}\Big( \sqrt{3} \cdotp\sqrt{10+2\sqrt{5} } + \sqrt{6-2\sqrt{5} }\Big)

\cos 42° =\cos(60° -18° ) = \cos 60°\cos 18°+\sin 60° \sin 18°

=\dfrac{1}{8}\Big(\sqrt{10+2\sqrt{5} } + \sqrt{3} \cdotp\sqrt{6-2\sqrt{5} }\Big)

Using the triple identity for sines function, we get 3 solutions for the cubic equation for \sin 14°.

x_0=\dfrac{ \sqrt{-\sin 42°+i\cos 42° } +\sqrt{-\sin 42°-i\cos 42°}}{2}

x_1=\dfrac{ ω\sqrt{-\sin 42°+i\cos 42° } +\overline{ω}\sqrt{-\sin 42°-i\cos42°}}{2}

x_2=\dfrac{ \overline{ω} \sqrt{-\sin 42°+i\cos 42° } +ω\sqrt{-\sin 42°-i\cos 42°}}{2}

To identify which angles they represent, we rewrite the solution according to equation (4). Then,

x = =\dfrac{1}{2}\Big( \sqrt{-\sin 42°+i\cos 42° }+\dfrac{1}{ \sqrt{-\sin 42°+i\cos 42° }} \Big)

=\dfrac{1}{2}\Big( \sqrt{\cos 132°+i\sin 132°}+\dfrac{1}{ \sqrt{\cos 132°+i\sin 132°}} \Big)

=\dfrac{1}{2}\Big( \cos(44°+ 120°k)+i\sin(44°+120k) +\dfrac{1}{\cos(44°+ 120°k)+i\sin(44°+120k) } \Big)

If k=0, x_0 = \cos 44°=\sin46°

If k = 1, x_1 = \cos 164° = -\sin 74°

If k=2, x_2 =\cos 284° =\sin 14°

It's found the third solution representing the value of \sin 14°, the first solution representing the value of \sin 46° and the second representing the negative value of \sin 74°

Similarly, establish a cubic equation for solving \cos 14° using the triple identity for cosines function which its cube roots given below,

x_0=\dfrac{1}{2}(\sqrt{\cos 42°+i\sin 42°}+\sqrt{\cos 42°-i\sin 42°})

x_1=\dfrac{1}{2}(ω\sqrt{\cos 42°+i\sin 42°}+\overline{ω}\sqrt{\cos 42°-i\sin 42°})

x_2=\dfrac{1}{2}(\overline{ω}\sqrt{\cos 42°+i\sin 42°}+ω\sqrt{\cos 42°-i\sin 42°})

Rewrite the solution according to the equation (4) to determine which angle they represent.

x=\dfrac{1}{2}\Big( \sqrt{\cos 42°+i\sin 42°}+\dfrac{1}{ \sqrt{\cos 42°+i\sin 42°}} \Big)

=\dfrac{1}{2}\Big( \cos(14°+ 120°k)+i\sin(14°+120k) +\dfrac{1}{\cos(14°+ 120°k)+i\sin(14°+120k) } \Big)

If k=0, x_0 = \cos 14°

If k = 1, x_1 = \cos 134° = -\cos 46°

If k=2, x_2 =\cos 254° =-\cos 74°

It shows that the first root represent the value of \cos 14°. The other two are negative numbers of \cos 46° and \cos 74°

## 16°

First determine the trig values for 48°.

\sin18\degree =\dfrac{-1+\sqrt{5} }{4} = \dfrac{\sqrt{6-2\sqrt{5} } }{4} ,

\cos 18\degree =\dfrac{\sqrt{10+2\sqrt{5} } }{4}

Given sines and cosines values for the special angles of 30° and 18°, the trig values for 48° could be computed by using sum identities.

\sin 48° =\sin(30°+18° ) = \sin 30°\cos 18°+\cos 30° \sin 18°

=\dfrac{1}{8}\Big(\sqrt{10+2\sqrt{5} } + \sqrt{3} \cdotp \sqrt{6-2\sqrt{5} }\Big)

\cos 48° =\cos(30°+18° ) = \cos 30°\cos 18°-\sin 30° \sin 18°

=\dfrac{1}{8}\Big( \sqrt{3} \cdotp \sqrt{10+2\sqrt{5} } -\sqrt{6-2\sqrt{5} }\Big)

Using the triple identity for sines function, we get 3 solutions for the cubic equation for \sin 16°.

x_0=\dfrac{ \sqrt{-\sin 48°+i\cos 48° } +\sqrt{-\sin 48°-i\cos 48°}}{2}

x_1=\dfrac{ ω\sqrt{-\sin 48°+i\cos 48° } +\overline{ω}\sqrt{-\sin 48°-i\cos48°}}{2}

x_2=\dfrac{ \overline{ω} \sqrt{-\sin 48°+i\cos 48° } +ω\sqrt{-\sin 42°-i\cos 48°}}{2}

To identify which angles they represent, we rewrite the solution according to equation (4). Then,

x = =\dfrac{1}{2}\Big( \sqrt{-\sin 48°+i\cos 48° }+\dfrac{1}{ \sqrt{-\sin 48°+i\cos 48° }} \Big)

=\dfrac{1}{2}\Big( \sqrt{\cos 138°+i\sin 138°}+\dfrac{1}{ \sqrt{\cos 138°+i\sin 138°}} \Big)

=\dfrac{1}{2}\Big( \cos(46°+ 120°k)+i\sin(46°+120k) +\dfrac{1}{\cos(46°+ 120°k)+i\sin(46°+120k) } \Big)

If k=0, x_0 = \cos 46°=\sin44°

If k = 1, x_1 = \cos 166° = -\sin 76°

If k=2, x_2 =\cos 286° =\sin 16°

It's found the third solution representing the value of \sin 16°, the first solution representing the value of \sin 44° and the second representing the negative value of \sin 76°

Similarly, establish a cubic equation for solving \cos 16° using the triple identity for cosines function which its cube roots given below,

x_0=\dfrac{1}{2}(\sqrt{\cos 48°+i\sin 48°}+\sqrt{\cos 48°-i\sin 48°})

x_1=\dfrac{1}{2}(ω\sqrt{\cos 48°+i\sin 48°}+\overline{ω}\sqrt{\cos 48°-i\sin 48°})

x_2=\dfrac{1}{2}(\overline{ω}\sqrt{\cos 48°+i\sin 42°}+ω\sqrt{\cos 48°-i\sin 48°})

Rewrite the solution according to the equation (4) to determine which angle they represent.

x=\dfrac{1}{2}\Big( \sqrt{\cos 48°+i\sin 48°}+\dfrac{1}{ \sqrt{\cos 48°+i\sin 48°}} \Big)

=\dfrac{1}{2}\Big( \cos(16°+ 120°k)+i\sin(16°+120k) +\dfrac{1}{\cos(16°+ 120°k)+i\sin(16°+120k) } \Big)

If k=0, x_0 = \cos 16°

If k = 1, x_1 = \cos 136° = -\cos 44°

If k=2, x_2 =\cos 256° =-\cos 76°

It shows that the first root represent the value of \cos 16°. The other two are negative numbers of \cos 44° and \cos 76°

## 17°

First determine the trig values for 51°.

\cos 36°= \dfrac{1+\sqrt{5} }{4}, \sin 36° =\dfrac{1}{4}\cdotp \sqrt{10-2\sqrt{5} }

\sin 15° =\dfrac{1}{2}\cdotp\sqrt{2-\sqrt{3} } =\dfrac{\sqrt{6}-\sqrt{2} }{4}

\cos 15° =\dfrac{1}{2}\cdotp\sqrt{2+\sqrt{3} } =\dfrac{\sqrt{6}+\sqrt{2} }{4}

Given sines and cosines values for the special angles of 54° and 15°, the trig values for 39° could be computed by using sum identities.

\sin 51° =\sin(36°+15° ) = \sin 36°\cos 15°+\cos 36° \sin 15°

=\dfrac{1}{8}\Big[ \sqrt{10-2\sqrt{5} } \cdotp\sqrt{2+\sqrt{3} } +(1+\sqrt{5}) \cdotp\sqrt{2-\sqrt{3} } \Big]

\cos 51° =\cos(36°+15° ) = \cos 36°\cos 15°-\sin 36° \sin 15°

=\dfrac{1}{8}\Big[ (1+\sqrt{5} ) \cdotp \sqrt{2+\sqrt{3} } -\sqrt{10-2\sqrt{5} } \cdotp\sqrt{2-\sqrt{3} } \Big]

Using the triple identity for sines function, we get 3 solutions for the cubic equation for \sin 16°.

x_0=\dfrac{ \sqrt{-\sin 51°+i\cos 51° } +\sqrt{-\sin 51°-i\cos 51°}}{2}

x_1=\dfrac{ ω\sqrt{-\sin 51°+i\cos 51° } +\overline{ω}\sqrt{-\sin 51°-i\cos51°}}{2}

x_2=\dfrac{ \overline{ω} \sqrt{-\sin 51°+i\cos 51° } +ω\sqrt{-\sin 51°-i\cos 51°}}{2}

To identify which angles they represent, we rewrite the solution according to equation (4). Then,

x = =\dfrac{1}{2}\Big( \sqrt{-\sin 51°+i\cos 51° }+\dfrac{1}{ \sqrt{-\sin 51°+i\cos 51° }} \Big)

=\dfrac{1}{2}\Big( \sqrt{\cos 141°+i\sin 141°}+\dfrac{1}{ \sqrt{\cos 141°+i\sin 141°}} \Big)

=\dfrac{1}{2}\Big( \cos(47°+ 120°k)+i\sin(47°+120k) +\dfrac{1}{\cos(47°+ 120°k)+i\sin(47°+120k) } \Big)

If k=0, x_0 = \cos 47°=\sin 43°

If k = 1, x_1 = \cos 167° = -\sin 77°

If k=2, x_2 =\cos 287° =\sin 17°

It's found the third solution representing the value of \sin 17°, the first solution representing the value of \sin 43° and the second representing the negative value of \sin 77°

Similarly, establish a cubic equation for solving \cos 17° using the triple identity for cosines function which its cube roots given below,

x_0=\dfrac{1}{2}(\sqrt{\cos 51°+i\sin 51°}+\sqrt{\cos 51°-i\sin 51°})

x_1=\dfrac{1}{2}(ω\sqrt{\cos 51°+i\sin 51°}+\overline{ω}\sqrt{\cos 51°-i\sin 51°})

x_2=\dfrac{1}{2}(\overline{ω}\sqrt{\cos 51°+i\sin 51°}+ω\sqrt{\cos 51°-i\sin 51°})

Rewrite the solution according to the equation (4) to determine which angle they represent.

x=\dfrac{1}{2}\Big( \sqrt{\cos 51°+i\sin 51°}+\dfrac{1}{ \sqrt{\cos 51°+i\sin 51°}} \Big)

=\dfrac{1}{2}\Big( \cos(17°+ 120°k)+i\sin(17°+120k) +\dfrac{1}{\cos(17°+ 120°k)+i\sin(17°+120k) } \Big)

If k=0, x_0 = \cos 17°

If k = 1, x_1 = \cos 137° = -\cos 43°

If k=2, x_2 =\cos 257° =-\cos 77°

It shows that the first root represent the value of \cos 17°. The other two are negative numbers of \cos 43° and \cos 77°

## 19°

First determine the trig values for 57°. We have determined the sines and cosines value for 33°. Using the co-function identities, we get

\$\cos 57° =\sin 33° =\dfrac{1}{8}\Big( \sqrt{6-2\sqrt{5} }\cdotp \sqrt{2+\sqrt{3} }+\sqrt{10+2\sqrt{5} }\cdotp \sqrt{2-\sqrt{3} }\Big)

\sin 57° =\cos 33° =\dfrac{1}{8}\Big( \sqrt{10+2\sqrt{5} }\cdotp \sqrt{2-\sqrt{3} }+ \sqrt{6-2\sqrt{5} }\cdotp\sqrt{2+\sqrt{3} }\Big)

Using the triple identity for sines function, we get 3 solutions for the cubic equation for \sin 16°.

x_0=\dfrac{ \sqrt{-\sin 57°+i\cos 57° } +\sqrt{-\sin 57°-i\cos 57°}}{2}

x_1=\dfrac{ ω\sqrt{-\sin 57°+i\cos 57° } +\overline{ω}\sqrt{-\sin 57°-i\cos57°}}{2}

x_2=\dfrac{ \overline{ω} \sqrt{-\sin 57°+i\cos 57° } +ω\sqrt{-\sin 57°-i\cos 57°}}{2}

To identify which angles they represent, we rewrite the solution according to equation (4). Then,

x = =\dfrac{1}{2}\Big( \sqrt{-\sin 57°+i\cos 57° }+\dfrac{1}{ \sqrt{-\sin 57°+i\cos 57° }} \Big)

=\dfrac{1}{2}\Big( \sqrt{\cos 147°+i\sin 147°}+\dfrac{1}{ \sqrt{\cos 147°+i\sin 147°}} \Big)

=\dfrac{1}{2}\Big( \cos(49°+ 120°k)+i\sin(49°+120k) +\dfrac{1}{\cos(49°+ 120°k)+i\sin(49°+120k) } \Big)

If k=0, x_0 = \cos 49°=\sin 41°

If k = 1, x_1 = \cos 169° = -\sin 79°

If k=2, x_2 =\cos 289° =\sin 19°

It's found the third solution representing the value of \sin 19°, the first solution representing the value of \sin 41° and the second representing the negative value of \sin 79°

Similarly, establish a cubic equation for solving \cos 19° using the triple identity for cosines function which its cube roots given below,

x_0=\dfrac{1}{2}(\sqrt{\cos 57°+i\sin 57°}+\sqrt{\cos 57°-i\sin 57°})

x_1=\dfrac{1}{2}(ω\sqrt{\cos 57°+i\sin 57°}+\overline{ω}\sqrt{\cos 57°-i\sin 57°})

x_2=\dfrac{1}{2}(\overline{ω}\sqrt{\cos 57°+i\sin 57°}+ω\sqrt{\cos 57°-i\sin 57°})

Rewrite the solution according to the equation (4) to determine which angle they represent.

x=\dfrac{1}{2}\Big( \sqrt{\cos 57°+i\sin 57°}+\dfrac{1}{ \sqrt{\cos 57°+i\sin 57°}} \Big)

=\dfrac{1}{2}\Big( \cos(19°+ 120°k)+i\sin(19°+120k) +\dfrac{1}{\cos(19°+ 120°k)+i\sin(19°+120k) } \Big)

If k=0, x_0 = \cos 19°

If k = 1, x_1 = \cos 139° = -\cos 41°

If k=2, x_2 =\cos 259° =-\cos 79°

It shows that the first root represent the value of \cos 19°. The other two are negative numbers of \cos 41° and \cos 79°

## 20° using triple identity

Using the triple identity for sines function, we get 3 solutions for the cubic equation.

x_0=\dfrac{ \sqrt{-\sin 60°+i\cos 60° } +\sqrt{-\sin 60°-i\cos 60°}}{2}

x_1=\dfrac{ ω \sqrt{-\sin 60°+i\cos 60° } +\overline{ω}\sqrt{-\sin 60°-i\cos60°}}{2}

x_2=\dfrac{\overline{ω} \sqrt{-\sin 60°+i\cos 60° } +ω\sqrt{-\sin 60°-i\cos 60°}}{2}

To identify which angles they represent, we rewrite the solution according to equation (4). Then,

x = =\dfrac{1}{2}\Big( \sqrt{-\sin 60°+i\cos 60° }+\dfrac{1}{ \sqrt{-\sin 60°+i\cos 60° }} \Big)

=\dfrac{1}{2}\Big( \sqrt{\cos 150°+i\sin 150°}+\dfrac{1}{ \sqrt{\cos 150°+i\sin 150°}} \Big)

=\dfrac{1}{2}\Big( \cos(50°+ 120°k)+i\sin(50°+120k) +\dfrac{1}{\cos(50°+ 120°k)+i\sin(50°+120k) } \Big)

If k=0, x_0 = \cos 50°=\sin40°

If k = 1, x_1 = \cos 170° = -\sin 80°

If k=2, x_2 =\cos 290° =\sin 20°

It's found the third solution representing the value of \sin 20°, the first solution representing the value of \sin 40° and the second representing the negative value of \sin 80°

Similarly, establish a cubic equation for solving \cos 20° using the triple identity for cosines function which its cube roots given below,

x_0=\dfrac{1}{2}(\sqrt{\cos 60°+i\sin 60°}+\sqrt{\cos 60°-i\sin 60°})

x_1=\dfrac{1}{2}(ω\sqrt{\cos 60°+i\sin 60°}+\overline{ω}\sqrt{\cos 60°-i\sin 60°})

x_2=\dfrac{1}{2}(\overline{ω}\sqrt{\cos 60°+i\sin 60°}+ω\sqrt{\cos 60°-i\sin 60°})

Rewrite the solution according to the equation (4) to determine which angle they represent.

x=\dfrac{1}{2}\Big( \sqrt{\cos 60°+i\sin 60°}+\dfrac{1}{ \sqrt{\cos 60°+i\sin 60°}} \Big)

=\dfrac{1}{2}\Big( \cos(20°+ 120°k)+i\sin(20°+120k) +\dfrac{1}{\cos(20°+ 120°k)+i\sin(20°+120k) } \Big)

If k=0, x_0 = \cos 20°

If k = 1, x_1 = \cos 140° = -\cos 40°

If k=2, x_2 =\cos 260° =-\cos 80°

It shows that the first root represent the value of \cos 20°. The other two are negative numbers of \cos 40° and \cos 80°

## 21°

First determine the trig values for 63°.

\cos 18° =\dfrac{\sqrt{10+2\sqrt{5} } }{4}

\sin 18° = \dfrac{-1+\sqrt{5} }{4}

Then, determine the trig values for 63° by using the sum identities.

\sin 63° =\sin(18°+45° ) = \sin 18°\cos 45°+\cos 18° \sin 45°

=\dfrac{1}{8}[(-\sqrt{2} +\sqrt{10} ) -(\sqrt{20+4\sqrt{5} }) ]

\cos 63° = \cos(18° +45° ) = \cos 18°\cos 45°-\sin 18° \sin 45°

=\dfrac{1}{8}[\sqrt{20+4\sqrt{5} } - (-\sqrt{2} +\sqrt{10}) ]

Using the triple identity for sines function, we get 3 solutions for the cubic equation for \sin 21°.

x_0=\dfrac{ \sqrt{-\sin 63°+i\cos 63° } +\sqrt{-\sin 63°-i\cos 63°}}{2}

x_1=\dfrac{ ω\sqrt{-\sin 63°+i\cos 63° } +\overline{ω}\sqrt{-\sin 63°-i\cos63°}}{2}

x_2=\dfrac{ \overline{ω} \sqrt{-\sin 63°+i\cos 63° } +ω\sqrt{-\sin 63°-i\cos 63°}}{2}

To identify which angles they represent, we rewrite the solution according to equation (4). Then,

x = =\dfrac{1}{2}\Big( \sqrt{-\cos 153°+i\sin 153°}+\dfrac{1}{ \sqrt{\cos 153°+i\sin 153°}} \Big)

=\dfrac{1}{2}\Big( \cos(51°+ 120°k)+i\sin(51°+120k) +\dfrac{1}{\cos(51°+ 120°k)+i\sin(51°+120k) } \Big)

If k=0, x_0 = \cos 51°=\sin 39°

If k = 1, x_1 = \cos 171° = -\sin 81°

If k=2, x_2 =\cos 291° =\sin 21°

It's found the third solution representing the value of \sin 21°, the first solution representing the value of \sin 39° and the second representing the negative value of \sin 81°

Similarly, establish a cubic equation for solving \cos 21° using the triple identity for cosines function which its cube roots given below,

x_0=\dfrac{1}{2}(\sqrt{\cos 63°+i\sin 63°}+\sqrt{\cos 63°-i\sin 63°})

x_1=\dfrac{1}{2}(ω\sqrt{\cos 63°+i\sin 63°}+\overline{ω}\sqrt{\cos 63°-i\sin 63°})

x_2=\dfrac{1}{2}(\overline{ω}\sqrt{\cos 63°+i\sin 63°}+ω\sqrt{\cos 63°-i\sin 63°})

Rewrite the solution according to the equation (4) to determine which angle they represent.

x=\dfrac{1}{2}\Big( \sqrt{\cos 63°+i\sin 63°}+\dfrac{1}{ \sqrt{\cos 63°+i\sin 63°}} \Big)

=\dfrac{1}{2}\Big( \cos(21°+ 120°k)+i\sin(21°+120k) +\dfrac{1}{\cos(21°+ 120°k)+i\sin(21°+120k) } \Big)

If k=0, x_0 = \cos 21°

If k = 1, x_1 = \cos 141° = -\cos 39°

If k=2, x_2 =\cos 261° =-\cos 81°

It shows that the first root represent the value of \cos 21°. The other two are negative numbers of \cos 39° and \cos 81°

## 22°

First determine the trig values for 66°.

\cos 36°= \dfrac{1+\sqrt{5} }{4}

\sin 36° =\dfrac{1}{4}\cdotp \sqrt{10-2\sqrt{5} }

Then, determine the trig values for 66° by using the sum identities.

\sin 66° =\sin(36°+30° ) = \sin 36°\cos 30°+\cos 36° \sin 30°

=\dfrac{1}{8}[\sqrt{3} \sqrt{10-2\sqrt{5} } -( 1+\sqrt{5}) ]

\cos 66° = \cos(36° +30° ) = \cos 36°\cos 30°-\sin 36° \sin 30°

=\dfrac{1}{8}[\sqrt{3} (1+\sqrt{5}) - \sqrt{10-2\sqrt{5} } ]

Using the triple identity for sines function, we get 3 solutions for the cubic equation for \sin 22°.

x_0=\dfrac{ \sqrt{-\sin 66°+i\cos 66° } +\sqrt{-\sin 66°-i\cos 66°}}{2}

x_1=\dfrac{ ω\sqrt{-\sin 66°+i\cos 66° } +\overline{ω}\sqrt{-\sin 66°-i\cos66°}}{2}

x_2=\dfrac{ \overline{ω} \sqrt{-\sin 66°+i\cos 66° } +ω\sqrt{-\sin 66°-i\cos 66°}}{2}

To identify which angles they represent, we rewrite the solution according to equation (4). Then,

x = =\dfrac{1}{2}\Big( \sqrt{-\cos 156°+i\sin 156°}+\dfrac{1}{ \sqrt{\cos 156°+i\sin 156°}} \Big)

=\dfrac{1}{2}\Big( \cos(52°+ 120°k)+i\sin(52°+120k) +\dfrac{1}{\cos(52°+ 120°k)+i\sin(52°+120k) } \Big)

If k=0, x_0 = \cos 52°=\sin 38°

If k = 1, x_1 = \cos 172° = -\sin 82°

If k=2, x_2 =\cos 292° =\sin 22°

It's found the third solution representing the value of \sin 22°, the first solution representing the value of \sin 38° and the second representing the negative value of \sin 82°

Similarly, establish a cubic equation for solving \cos 22° using the triple identity for cosines function which its cube roots given below,

x_0=\dfrac{1}{2}(\sqrt{\cos 66°+i\sin 66°}+\sqrt{\cos 66°-i\sin 66°})

x_1=\dfrac{1}{2}(ω\sqrt{\cos 66°+i\sin 66°}+\overline{ω}\sqrt{\cos 66°-i\sin 66°})

x_2=\dfrac{1}{2}(\overline{ω}\sqrt{\cos 66°+i\sin 66°}+ω\sqrt{\cos 66°-i\sin 66°})

Rewrite the solution according to the equation (4) to determine which angle they represent.

x=\dfrac{1}{2}\Big( \sqrt{\cos 66°+i\sin 66°}+\dfrac{1}{ \sqrt{\cos 66°+i\sin 66°}} \Big)

=\dfrac{1}{2}\Big( \cos(22°+ 120°k)+i\sin(22°+120k) +\dfrac{1}{\cos(22°+ 120°k)+i\sin(22°+120k) } \Big)

If k=0, x_0 = \cos 22°

If k = 1, x_1 = \cos 142° = -\cos 38°

If k=2, x_2 =\cos 262° =-\cos 82°

It shows that the first root represent the value of \cos 22°. The other two are negative numbers of \cos 38° and \cos 82°

## 23°

First determine the trig values for 69°.

\sin 54°= \dfrac{1+\sqrt{5} }{4}

\cos 54° =\dfrac{1}{4}\cdotp \sqrt{10-2\sqrt{5} }

\sin 15° =\dfrac{1}{2}\cdotp\sqrt{2-\sqrt{3} } =\dfrac{\sqrt{6}-\sqrt{2} }{4}

\cos 15° =\dfrac{1}{2}\cdotp\sqrt{2+\sqrt{3} } =\dfrac{\sqrt{6}+\sqrt{2} }{4}

Then, determine the trig values for 69° by using the sum identities.

\sin 69° =\sin(54°+15° ) = \sin 54°\cos 15°+\cos 54° \sin 15°

=\dfrac{1}{8}[(1+\sqrt{5} ) \sqrt{2+\sqrt{3} } + \sqrt{10-2\sqrt{5} } \sqrt{2-\sqrt{3} } ]

\cos 69° = \cos(54° +15° ) = \cos 54°\cos 15°-\sin 54° \sin 15°

=\dfrac{1}{8}[ \sqrt{10-2\sqrt{5} }(\sqrt{2+\sqrt{3} } - (1+\sqrt{5} )\sqrt{2-\sqrt{3} }]

Using the triple identity for sines function, we get 3 solutions for the cubic equation for \sin 23°.

x_0=\dfrac{ \sqrt{-\sin 69°+i\cos 69° } +\sqrt{-\sin 69°-i\cos 69°}}{2}

x_1=\dfrac{ ω\sqrt{-\sin 69°+i\cos 69° } +\overline{ω}\sqrt{-\sin 69°-i\cos69°}}{2}

x_2=\dfrac{ \overline{ω} \sqrt{-\sin 69°+i\cos 69° } +ω\sqrt{-\sin 69°-i\cos 69°}}{2}

To identify which angles they represent, we rewrite the solution according to equation (4). Then,

x = =\dfrac{1}{2}\Big( \sqrt{-\cos 159°+i\sin 159°}+\dfrac{1}{ \sqrt{\cos 159°+i\sin 159°}} \Big)

=\dfrac{1}{2}\Big( \cos(53°+ 120°k)+i\sin(53°+120k) +\dfrac{1}{\cos(53°+ 120°k)+i\sin(53°+120k) } \Big)

If k=0, x_0 = \cos 53°=\sin 37°

If k = 1, x_1 = \cos 173° = -\sin 83°

If k=2, x_2 =\cos 293° =\sin 23°

It's found the third solution representing the value of \sin 23°, the first solution representing the value of \sin 37° and the second representing the negative value of \sin 83°

Similarly, establish a cubic equation for solving \cos 23° using the triple identity for cosines function which its cube roots given below,

x_0=\dfrac{1}{2}(\sqrt{\cos 69°+i\sin 69°}+\sqrt{\cos 69°-i\sin 69°})

x_1=\dfrac{1}{2}(ω\sqrt{\cos 69°+i\sin 69°}+\overline{ω}\sqrt{\cos 69°-i\sin 69°})

x_2=\dfrac{1}{2}(\overline{ω}\sqrt{\cos 69°+i\sin 69°}+ω\sqrt{\cos 69°-i\sin 69°})

Rewrite the solution according to the equation (4) to determine which angle they represent.

x=\dfrac{1}{2}\Big( \sqrt{\cos 69°+i\sin 69°}+\dfrac{1}{ \sqrt{\cos 69°+i\sin 69°}} \Big)

=\dfrac{1}{2}\Big( \cos(23°+ 120°k)+i\sin(23°+120k) +\dfrac{1}{\cos(23°+ 120°k)+i\sin(23°+120k) } \Big)

If k=0, x_0 = \cos 23°

If k = 1, x_1 = \cos 143° = -\cos 37°

If k=2, x_2 =\cos 263° =-\cos 83°

It shows that the first root represent the value of \cos 23°. The other two are negative numbers of \cos 37° and \cos 83°

## 25°

Using the triple identity for sines function, we get 3 solutions for the cubic equation.

x_0=\dfrac{ \sqrt{-\sin 75°+i\cos 75° } +\sqrt{-\sin 75°-i\cos 75°}}{2}

x_1=\dfrac{ ω \sqrt{-\sin 75°+i\cos 75° } +\overline{ω}\sqrt{-\sin 75°-i\cos75°}}{2}

x_2=\dfrac{ \overline{ω}\sqrt{-\sin 75°+i\cos 75° } +ω\sqrt{-\sin 75°-i\cos 75°}}{2}

To identify which angles they represent, we rewrite the solution according to equation (4). Then,

x = =\dfrac{1}{2}\Big( \sqrt{-\sin 75°+i\cos 75° }+\dfrac{1}{ \sqrt{-\sin 75°+i\cos 75° }} \Big)

=\dfrac{1}{2}\Big( \sqrt{\cos 165°+i\sin 165°}+\dfrac{1}{ \sqrt{\cos 165°+i\sin 165°}} \Big)

=\dfrac{1}{2}\Big( \cos(55°+ 120°k)+i\sin(55°+120k) +\dfrac{1}{\cos(55°+ 120°k)+i\sin(55°+120k) } \Big)

If k=0, x_0 = \cos 55°=\sin35°

If k = 1, x_1 = \cos 175° = -\sin 85°

If k=2, x_2 =\cos 295° =\sin 25°

It's found the third solution representing the value of \sin 25°, the first solution representing the value of \sin 35° and the second representing the negative value of \sin 85°

Similarly, establish a cubic equation for solving \cos 25° using the triple identity for cosines function which its cube roots given below,

x_0=\dfrac{1}{2}(\sqrt{\cos 75°+i\sin 75°}+\sqrt{\cos 75°-i\sin 75°})

x_1=\dfrac{1}{2}(ω\sqrt{\cos 75°+i\sin 75°}+\overline{ω}\sqrt{\cos 75°-i\sin 75°})

x_2=\dfrac{1}{2}(\overline{ω}\sqrt{\cos 75°+i\sin 75°}+ω\sqrt{\cos 75°-i\sin 75°})

Rewrite the solution according to the equation (4) to determine which angle they represent.

x=\dfrac{1}{2}\Big( \sqrt{\cos 75°+i\sin 75°}+\dfrac{1}{ \sqrt{\cos 75°+i\sin 75°}} \Big)

=\dfrac{1}{2}\Big( \cos(25°+ 120°k)+i\sin(25°+120k) +\dfrac{1}{\cos(25°+ 120°k)+i\sin(25°+120k) } \Big)

If k=0, x_0 = \cos 25°

If k = 1, x_1 = \cos 145° = -\cos 35°

If k=2, x_2 =\cos 265° =-\cos 85°

It shows that the first root represent the value of \cos 25°. The other two are negative numbers of \cos 35° and \cos 85°

## 26°

78=60+18

First determine the trig values for 78°.

\cos 18° =\dfrac{\sqrt{10+2\sqrt{5} } }{4}

\sin 18° = \dfrac{-1+\sqrt{5} }{4}

Then, determine the trig values for 78° by using the sum identities.

\sin 78° =\sin(18°+60° ) = \sin 18°\cos 60°+\cos 18° \sin 60°

=\dfrac{1}{8}[(-\sqrt{1} +\sqrt{5} ) +\sqrt{3} ( \sqrt{10+2\sqrt{5} }) ]

\cos 78° = \cos(18° +60° ) = \cos 18°\cos 60°-\sin 18° \sin 60°

=\dfrac{1}{8}[ \sqrt{10+2\sqrt{5} } - \sqrt{3} (\sqrt{1} +\sqrt{5}) ]

Using the triple identity for sines function, we get 3 solutions for the cubic equation for \sin 26°.

x_0=\dfrac{ \sqrt{-\sin 78°+i\cos 78° } +\sqrt{-\sin 78°-i\cos 78°}}{2}

x_1=\dfrac{ ω\sqrt{-\sin 78°+i\cos 78° } +\overline{ω}\sqrt{-\sin 78°-i\cos78°}}{2}

x_2=\dfrac{ \overline{ω} \sqrt{-\sin 78°+i\cos 78° } +ω\sqrt{-\sin 78°-i\cos 78°}}{2}

To identify which angles they represent, we rewrite the solution according to equation (4). Then,

x = =\dfrac{1}{2}\Big( \sqrt{-\cos 168°+i\sin 153°}+\dfrac{1}{ \sqrt{\cos 168°+i\sin 168°}} \Big)

=\dfrac{1}{2}\Big( \cos(56°+ 120°k)+i\sin(56°+120k) +\dfrac{1}{\cos(56°+ 120°k)+i\sin(56°+120k) } \Big)

If k=0, x_0 = \cos 56°=\sin 34°

If k = 1, x_1 = \cos 176° = -\sin 86°

If k=2, x_2 =\cos 296° =\sin 26°

It's found the third solution representing the value of \sin 26°, the first solution representing the value of \sin 34° and the second representing the negative value of \sin 86°

Similarly, establish a cubic equation for solving \cos 26° using the triple identity for cosines function which its cube roots given below,

x_0=\dfrac{1}{2}(\sqrt{\cos 78°+i\sin 78°}+\sqrt{\cos 78°-i\sin 78°})

x_1=\dfrac{1}{2}(ω\sqrt{\cos 78°+i\sin 78°}+\overline{ω}\sqrt{\cos 78°-i\sin 78°})

x_2=\dfrac{1}{2}(\overline{ω}\sqrt{\cos 78°+i\sin 78°}+ω\sqrt{\cos 78°-i\sin 78°})

Rewrite the solution according to the equation (4) to determine which angle they represent.

x=\dfrac{1}{2}\Big( \sqrt{\cos 78°+i\sin 78°}+\dfrac{1}{ \sqrt{\cos 78°+i\sin 78°}} \Big)

=\dfrac{1}{2}\Big( \cos(26°+ 120°k)+i\sin(26°+120k) +\dfrac{1}{\cos(26°+ 120°k)+i\sin(26°+120k) } \Big)

If k=0, x_0 = \cos 26°

If k = 1, x_1 = \cos 146° = -\cos 34°

If k=2, x_2 =\cos 266° =-\cos 86°

It shows that the first root represent the value of \cos 26°. The other two are negative numbers of \cos 34° and \cos 86°

## 28°

First determine the trig values for 84°.

\sin 54°= \dfrac{1+\sqrt{5} }{4}

\cos 54° =\dfrac{1}{4}\cdotp \sqrt{10-2\sqrt{5} }

Then, determine the trig values for 84° by using the sum identities.

\sin 84° =\sin(54°+30° ) = \sin 54°\cos 30°+\cos 54° \sin 30°

=\dfrac{1}{8}[(1+\sqrt{5} ) \sqrt{3} + \sqrt{10-2\sqrt{5} } ]

\cos 84° = \cos(54° +30° ) = \cos 54°\cos 30°-\sin 54° \sin 30°

=\dfrac{1}{8}[ \sqrt{3} \sqrt{10-2\sqrt{5} } - (1+\sqrt{5} )]

Using the triple identity for sines function, we get 3 solutions for the cubic equation for \sin 28°.

x_0=\dfrac{ \sqrt{-\sin 84°+i\cos 84° } +\sqrt{-\sin 84°-i\cos 84°}}{2}

x_1=\dfrac{ ω\sqrt{-\sin 84°+i\cos 84° } +\overline{ω}\sqrt{-\sin 84°-i\cos84°}}{2}

x_2=\dfrac{ \overline{ω} \sqrt{-\sin 84°+i\cos 84° } +ω\sqrt{-\sin 84°-i\cos 84°}}{2}

To identify which angles they represent, we rewrite the solution according to equation (4). Then,

x = =\dfrac{1}{2}\Big( \sqrt{-\cos 174°+i\sin 174°}+\dfrac{1}{ \sqrt{\cos 174°+i\sin 174°}} \Big)

=\dfrac{1}{2}\Big( \cos(58°+ 120°k)+i\sin(58°+120k) +\dfrac{1}{\cos(58°+ 120°k)+i\sin(58°+120k) } \Big)

If k=0, x_0 = \cos 58°=\sin 32°

If k = 1, x_1 = \cos 178° = -\sin 88°

If k=2, x_2 =\cos 298° =\sin 28°

It's found the third solution representing the value of \sin 28°, the first solution representing the value of \sin 32° and the second representing the negative value of \sin 88°

Similarly, establish a cubic equation for solving \cos 28° using the triple identity for cosines function which its cube roots given below,

x_0=\dfrac{1}{2}(\sqrt{\cos 84°+i\sin 84°}+\sqrt{\cos 84°-i\sin 84°})

x_1=\dfrac{1}{2}(ω\sqrt{\cos 84°+i\sin 84°}+\overline{ω}\sqrt{\cos 84°-i\sin 84°})

x_2=\dfrac{1}{2}(\overline{ω}\sqrt{\cos 84°+i\sin 84°}+ω\sqrt{\cos 84°-i\sin 84°})

Rewrite the solution according to the equation (4) to determine which angle they represent.

x=\dfrac{1}{2}\Big( \sqrt{\cos 84°+i\sin 84°}+\dfrac{1}{ \sqrt{\cos 84°+i\sin 84°}} \Big)

=\dfrac{1}{2}\Big( \cos(28°+ 120°k)+i\sin(28°+120k) +\dfrac{1}{\cos(28°+ 120°k)+i\sin(28°+120k) } \Big)

If k=0, x_0 = \cos 28°

If k = 1, x_1 = \cos 148° = -\cos 32°

If k=2, x_2 =\cos 268° =-\cos 88°

It shows that the first root represent the value of \cos 23°. The other two are negative numbers of \cos 37° and \cos 88°

## 29°

First determine the trig values for 87°.

\sin 15° =\dfrac{1}{2}\cdotp\sqrt{2-\sqrt{3} } =\dfrac{\sqrt{6}-\sqrt{2} }{4}

\cos 15° =\dfrac{1}{2}\cdotp\sqrt{2+\sqrt{3} } =\dfrac{\sqrt{6}+\sqrt{2} }{4}

\cos 72\degree =\dfrac{-1+\sqrt{5} }{4} = \dfrac{\sqrt{6-2\sqrt{5} } }{4} ,

\sin 72\degree =\dfrac{\sqrt{10+2\sqrt{5} } }{4}

Given sines and cosines values for the special angles of 15° and 72°, the trig values for 87° could be computed by using sum identities.

\sin 87° =\sin(72° +15° ) = \sin 72°\cos 15° +\cos 72° \sin 15°

=\dfrac{1}{8}\Big( \sqrt{10+2\sqrt{5} } \cdotp \sqrt{2+\sqrt{3} }+ \sqrt{6-2\sqrt{5} }\cdotp \sqrt{2-\sqrt{3} }\Big)

\cos 87° =\cos(72° +15° ) = \cos 72°\cos 15° -\sin 72° \sin 15°

=\dfrac{1}{8}\Big( \sqrt{6-2\sqrt{5} }\cdotp \sqrt{2-\sqrt{3} }+ \sqrt{10+2\sqrt{5} } \cdotp\sqrt{2+\sqrt{3} }\Big)

Using the triple identity for sines function, we get 3 solutions for the cubic equation for \sin 29°.

x_0=\dfrac{ \sqrt{-\sin 87°+i\cos 87° } +\sqrt{-\sin 87°-i\cos 87°}}{2}

x_1=\dfrac{ ω\sqrt{-\sin 87°+i\cos 87° } +\overline{ω}\sqrt{-\sin 87°-i\cos87°}}{2}

x_2=\dfrac{ \overline{ω} \sqrt{-\sin 87°+i\cos 87° } +ω\sqrt{-\sin 87°-i\cos 87°}}{2}

To identify which angles they represent, we rewrite the solution according to equation (4). Then,

x = =\dfrac{1}{2}\Big( \sqrt{-\sin 87°+i\cos 87° }+\dfrac{1}{ \sqrt{-\sin 87°+i\cos 87° }} \Big)

=\dfrac{1}{2}\Big( \sqrt{\cos 177°+i\sin 177°}+\dfrac{1}{ \sqrt{\cos 177°+i\sin 177°}} \Big)

=\dfrac{1}{2}\Big( \cos(59°+ 120°k)+i\sin(59°+120k) +\dfrac{1}{\cos(59°+ 120°k)+i\sin(59°+120k) } \Big)

If k=0, x_0 = \cos 59°=\sin 31°

If k = 1, x_1 = \cos 179° = -\sin 89°

If k=2, x_2 =\cos 299° =\sin 29°

It's found the third solution representing the value of \sin 29°, the first solution representing the value of \sin 31° and the second representing the negative value of \sin 89°

Similarly, establish a cubic equation for solving \cos 29° using the triple identity for cosines function which its cube roots given below,

x_0=\dfrac{1}{2}(\sqrt{\cos 87°+i\sin 87°}+\sqrt{\cos 87°-i\sin 87°})

x_1=\dfrac{1}{2}(ω\sqrt{\cos 87°+i\sin 87°}+\overline{ω}\sqrt{\cos 87°-i\sin 87°})

x_2=\dfrac{1}{2}(\overline{ω}\sqrt{\cos 87°+i\sin 87°}+ω\sqrt{\cos 87°-i\sin 87°})

Rewrite the solution according to the equation (4) to determine which angle they represent.

x=\dfrac{1}{2}\Big( \sqrt{\cos 87°+i\sin 87°}+\dfrac{1}{ \sqrt{\cos 87°+i\sin 87°}} \Big)

=\dfrac{1}{2}\Big( \cos(29°+ 120°k)+i\sin(29°+120k) +\dfrac{1}{\cos(29°+ 120°k)+i\sin(29°+120k) } \Big)

If k=0, x_0 = \cos 29°

If k = 1, x_1 = \cos 149° = -\cos 31°

If k=2, x_2 =\cos 269° =-\cos 89°

It shows that the first root represent the value of \cos 29°. The other two are negative numbers of \cos 31° and \cos 89°

## Summary

Trigonometric values of all integer angles in 90° could be expressed in the expressions of real numbers or complex numbers. The computation starts from special angles whose trigonometric values could be obtained geometrically. By using sum of two or three angles identities, co-function identities, double or half identities, the exact trig value of all angles that are multiples of 3 could be computed to an expression of real number. For others that are not multiples of 3, a cubic equation could be established for solutions given that the triple angle of the variable is available. However, there's only one out of three solutions that represents the trig value of the variable. The other two solutions represent other two angles in the complex plane that satisfies the cubic equation.

Collected in the board: Trigonometry

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