﻿ 3 Methods for Derivation of the cosine difference formula

# 3 Methods for Derivation of the cosine difference formula

The cosine difference formula is one of the important trigonometric identities. Once this identity is established it can be used to easily derive other important identities. This article is to derive the formula using multiple geometric methods.

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

## Method 1 - using distance formula in a unit circle

First let's observe two central angles in a circle whose radius is defined as 1. \alpha and \beta share the same vertex and a common arm, which coincides with axis x in the right direction. The other arms of \alpha and \beta intersect the circle at the points (x_1,y_1) and (x_2,y_2). As shown in the figure, the chord length subtended by the angle \alpha - \beta is determined by coordinates of the two intersection points. By using distance formula, the length of the chord is given as

\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}

Next let's observe a unit circle with a central angle whose measure is equal to \alpha - \beta and with one of its arms coinciding with axis x. Another arm of the angle intersects the circle at the point (m,n). Using the distance formula, the length of the chord determined by the point (m,n) and (1,0) can be calculated as,

\sqrt{(m-1)^2+n^2}

Since the chords in the two unit circle subtend central angles of the same measure, they have the same length. Then the following equation is established,

\sqrt{(x_1-x_2)^2+(y_1-y_2)^2} =\sqrt{(m-1)^2+n^2}
(1)

Square the both sides of the equation

(x_1-x_2)^2+(y_1-y_2)^2 = (m-1)^2+n^2

x^2_1+x^2_2-2x_1x_2+y^2_1+y^2_2-2y_1y_2=m^2-2m+1+n^2
(2)

Since any points on a unit circle to the origin has the same distance 1,

x^2_1+y^2_1 = 1, x^2_2+y^2_2 = 1, m^2+n^2=1

Equation (2) is simplified as,

2- 2x_1x_2-2y_1y_2 = 2-2m

x_1x_2+y_1y_2 = m
(3)

Since the radius of a unit circle is 1, it's obvious that

x_1 = \cos \alpha, x_2 = \cos \beta, y_1 = \sin \alpha, y_2 = \sin \beta

and

m = \cos (\alpha -\beta )

Therefore,

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

## Method 2 - using Law of Cosine and Pythagorean Theorem

In the unit circle shown in the figure, the central angle \alpha and \beta share a common arm which coincides with axis x in the right direction. The other arms of the two angles intersect the circle at points B and A.

Construct perpendicular segments from B to axis x and y with intersection points C and D.

Similarly, construct perpendicular segments from A to axis x and y with intersection points E and F.

Apply Law of Cosine in ther triangle ABO ,

\begin{aligned} AB^2&=OA^2+OB^2 - 2OA\cdot OB\cdot\cos(\alpha -\beta )\\ &= 1+1 -2\cdot1\cdot1\cdot\cos(\alpha -\beta ) \\ &= 2-2\cos(\alpha -\beta ) \\ \end{aligned}

(3)

Extend the altitude AE which intersects BC at point G.

Since \triangle AGB is a right triangle, according to Pythagorean Theorem,

\begin{aligned} AB^2 &= BG^2 + AG^2 \\ &= DE^2 +CF^2 \\ &= (OD-OE)^2 + (OC+OF)^2 \\ &= (OB\sin \alpha -OA \sin \beta )^ 2 + (-OB\cos\alpha +OA\cos\beta)^2 \\ &= (\sin \alpha -\sin \beta )^ 2 +(\cos\beta-\cos\alpha )^2 \\ &= 2-2\sin\alpha\cos\beta - 2\cos\alpha \sin\beta \\ \end{aligned}

(4)

Combine the expression (3), (4) and the cosine difference formula is verified.

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

## Method 3 - using Distance formula and Law of Cosine

This method is similar to method 1 in that it also uses distance formula. However, the steps are simplified by using Law of Cosine.

In the figure, the central angle \alpha and \beta start from the same arm which coincides with axis x of a unit circle and intersects the circle at point C. The other arms of them intersect the circle at point A and B.

So the coordinates of A, B are obtained as,

A (\cos \alpha, \sin \alpha )

B (\cos \beta , \sin \beta )

The length of the chord AB is defined by the coordinates of point A and B. Using the distance formula,

AB^2 = (\cos\alpha- \cos\beta)^2 + ( \sin\alpha-\sin\beta)^2

AB^2 = 2-2(\cos\alpha\cos\beta - \sin\alpha\sin\beta)
(5)

In the meantime, \triangle AOB is a isosceles triangle with equal measure of sides AO and BO, which is the dadius of the unit circle.

AO=BO =1

The length of the third side AB is determined by its subtended angle \alpha -\beta. Using the Law of Cosine,

\begin{aligned} AB^2&=OA^2+OB^2 - 2OA\cdot OB\cdot\cos(\alpha -\beta )\\ &= 1+1 -2\cdot1\cdot1\cdot\cos(\alpha -\beta ) \\ &= 2-2\cos(\alpha -\beta ) \\ \end{aligned}

Combine with expression (5), again we have proved the validation of the cosine difference formula.

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

Collected in the board: Trigonometry

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