﻿ Derivation of trigonometric Triple angle identities

# Derivation of trigonometric Triple angle identities

Triple angle identities give relationship between trigonometric functions of an angle and the one that is three multiples of the angle. Triple angle identities can also expressed to transform an trig function of power of three to trigonometric functions of lower power.

There are 6 triple angle identities corresponding to 6 elementary trigonometric functions. However, the most frequent used ones are for sines, cosine and tangent functions. In order to derive the triple angle identities, it is the premise to recognize the fundamental trigonometric identities such as double angle, sum and product identities.

## Triple angle identity for sine function

First, convert the triple angle to the sum of double angle and single angle. And then, use the Sum identity for sine function, double angle identities and Pythagorean identity for derivation.

\begin{aligned} \sin3α &=\sin(2α+α) \\ &=\sin2α \cos α+\cos2α \sin α \\ &=2\sin α \cos α\cos α+(1-2\sin^2α) \sin α \\ &=2\sin α(1-\sin^2α)+(1-2\sin^2α)\sin α \\ &=3\sin α-4\sin^3α \end{aligned}

Hence, triple angle identity for sine function looks a form of a depressed cubic function.

\sin3α = 3\sin α-4\sin^3α
(1)

Noticed the special numbers of coefficients of terms on right hand side, the triple angle identity could be transformed to the interesting form.

\begin{aligned} \sin 3α &=3\sin α-4\sin^3α \\&=4\sin α(\dfrac{3}{4} -\sin^2α) \\&=4\sin α[(\dfrac{\sqrt{3} }{2})^2-\sin^2α] \\&=4\sin α(\sin^2 60°-\sin^2α) \\&=4\sin α(\sin60°+\sin α)(\sin60°-\sin α) \\&=4\sin α\cdot2\sin\dfrac{60+α}{2} \cos\dfrac{60-α}{2}\cdot2\sin\dfrac{60-α}{2}\cos\dfrac{60-α}{2} \\&=4\sin α\sin(60°+α)\sin(60°-α) \end{aligned}

The triple identity could be transformed to the form of product of sine function of single angle and two symmetric angles about 60°.

\sin 3α = 4\sin α\sin(60°+α)\sin(60°-α)
(2)

## Triple angle identity for cosine function

Proof of triple angle identity for cosine function is similar to the steps for sine function.

\begin{aligned} \cos3α&=\cos(2α+α) \\ &=\cos 2α\cos α-\sin2α \sinα \\&=(2\cos^2α-1)\cos α-2(1-\cos^2α)\cos α \\&=4\cos^3α-3\cos α \end{aligned}

Hence, triple angle identity for cosine function is a form of a depressed cubic function in terms of cosine of single angle.

\cos3α = 4\cos^3α-3\cos α
(3)

In the same manner, the triple function for cosine function could be written as the product of cosine of single angle and two symmetric angles.

\begin{aligned} \cos3α&=4\cos³α-3\cos α \\&=4\cos α(\cos^2α-\dfrac{3}{4} ) \\&=4\cos α[\cos^2α-(\dfrac{\sqrt{3} }{2} )^2] \\&=4\cos α(\cos^2α-\cos^2 30°) \\&=4\cos α(\cos α+\cos30°)(\cos α-\cos30°) \\&=4\cos α \cdot 2\cos\dfrac{30+α}{2} \cos\dfrac{30-α}{2} \cdot (-2\sin\dfrac{30+α}{2}\sin\dfrac{30-α}{2}) \\&=-4\cos α\sin(α+30°)\sin(α-30°) \\&=-4\cos α\sin[90°-(60°-a)]\sin[-90°+(60°+α)] \\&=-4\cos α\cos(60°-α)[-\cos(60°+α)] \\&=4\cos α\cos(60°-α)\cos(60°+α) \end{aligned}

that is,

\cos3α = 4\cos α\cos(60°-α)\cos(60°+α)
(4)

Rewriting equation (1) and (3) gives the form of power reducing

for sine function

\sin^3α = \dfrac{3\sin α- \sin3α }{4}
(5)

and

for cosine function

\cos^3α = \dfrac{\cos3α+3\cos α}{4}
(6)

## Triple angle identity for tangent function

Triple angle identity for tangent function could be obtained by dividing sine by cosine identity in their product form.

Dividing (2) by (4) yields

\tan3α=\tan α\tan(60°-α)\tan(60°+α)
(7)

Using the sum and difference identities for tangent functions

\tan(\alpha+\beta) = \dfrac{\tan \alpha + \tan \beta }{1-\tan \alpha\tan \beta}
\tan(\alpha -\beta) = \dfrac{\tan \alpha - \tan \beta }{1+\tan \alpha\tan \beta}

\tan3α could be represented as the expression in terms of \tan α.

\begin{aligned} \tan3α&=\tan α\tan(60°-α)\tan(60°+α) \\&=\tan α \dfrac{\tan 60°- \tan α }{1+\tan 60°\tan α} \cdot \dfrac{\tan 60°+ \tan α }{1-\tan 60°\tan α} \\&=\tan α \dfrac{\sqrt{3} - \tan α }{1+\sqrt{3} \tan α} \cdot \dfrac{\sqrt{3} + \tan α }{1-\sqrt{3} \tan α} \\ &=\dfrac{3\tanα-\tan^3α}{1-3\tanα } \end{aligned}

that is

\tan3α=\dfrac{3\tanα-\tan^3α}{1-3\tanα }
(8)

Collected in the board: Trigonometry

Steven Zheng posted 5 months ago

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