Derivation of the cosine sum identity using geometric methods

There are multiple methods for derivation of cosine difference identity. Once this identity is present, it's easy to derive other identities. For example, the cosine sum identity is another frequently used identities to solve trigonometric problems. Given the cosine difference identity,

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

Let \beta be the minus of \theta, the above equation is converted to,

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

Using symmetry identities, the above expression is transformed to the identity for cosine of two angles addition,

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

What about any method using geometric shapes for proof of cosine sum identity? We will explore the methods in this article.

Using distance formula

In the figure is a circle whose radius is of measure of 1.

Angle \alpha and angle -\beta share a common arm, which coincides with axis x in the right direction. A, B, D are intersection points of their arms with the unit circle.


Angle \beta shares the same arm with the other arm of \alpha and intersects the circle at point C on the other arm.


Since OA,OB, OC, OD are the radius of the circle with measure of 1, the coordinates of point A,B, C, D are as follow,


A (1,0),

B (\cos\alpha, \sin\alpha),

C(\cos(\alpha+\beta),\sin(\alpha+\beta) ),

D (\cos\beta,-\sin \beta )

Using the distance formula, the length of chords AC and RD is expressed below,


AC^2 = (\cos(\alpha+\beta)-1)^2 +\sin^2(\alpha+\beta) = 2-2\cos(\alpha+\beta)
(1)


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


BD^2 =2 - 2(\cos\alpha\cos\beta- \sin\alpha\sin \beta)
(2)

Since chords AC and RD subtend the central angle of the same measure \alpha + \beta


AC = BD


Expression (1) is equal to expression (2)


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


Simplify the equation, we get the cosine sum identity.


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

Using right triangles


In the figure,

\alpha = \angle BAC

\beta = \angle CAD


Construct perpendicular DE to AB


Construct perpendicular DF to AC, FG to AB and FH to DE


Since \triangle AEI and \triangle DFI are right triangles with vertical angles,


\angle FDH = \alpha

\cos(\alpha +\beta ) = \dfrac{AE}{AD}


\because EG = HF


\therefore \dfrac{AE}{AD} = \dfrac{AG-EG}{AD} =\dfrac{AG-HF}{AD} = \dfrac{AG}{AD}-\dfrac{HF}{AD}


= \dfrac{AG}{AF}\cdotp \dfrac{AF}{AD} -\dfrac{HF}{DF}\cdotp \dfrac{DF}{AD}


\because \dfrac{AG}{AF}=\cos \alpha, \dfrac{AF}{AD} = \cos \beta, \dfrac{HF}{DF} = \sin \alpha, \dfrac{DF}{AD} =\sin \beta



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


This is the end of our proof for the cosine sum identity.

Collected in the board: Trigonometry

Steven Zheng Steven Zheng posted 2 months ago

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