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Q.
The value of $\cos (x+y)$ is equal to
Trigonometric Functions
Solution:
Consider the unit circle with centre at the origin. Let $x$ be the angle $P_4 O P_1$ and $y$ be the angle $P_1 O P_2$. Then, $(x+y)$ is the angle $P_4 O P_2$. Also, let $(-y)$ be the angle $P_4 O P_3$. Therefore, $P_1, P_2, P_3$ and $P_4$ will have the coordinates $P_1(\cos x, \sin x), P_2[\cos (x+y), \sin (x+y)]$,
$P_3[\cos (-y), \sin (-y)] \text { and } P_4(1,0) \text {. }$
Consider the triangles $P_1 O P_3$ and $P_2 O_4$. They are congruent. Therefore, $P_1 P_3$ and $P_2 P_4$ are equal. By using distance formula, we get
$P_1 P_3^2= {[\cos x-\cos (-y)]^2+[\sin x-\sin (-y)]^2 } $
$= (\cos x-\cos y)^2+(\sin x+\sin y)^2 $
$= \cos ^2 x+\cos ^2 y-2 \cos x \cos y+\sin ^2 x $
$ +\sin ^2 y+2 \sin x \sin y$
$ =2-2(\cos x \cos y-\sin x \sin y) $
Also, $ P_2 P_4^2=[1-\cos (x+y)]^2+[0-\sin (x+y)]^2 $
$ =1-2 \cos (x+y)+\cos ^2(x+y)+\sin ^2(x+y)$
$ =2-2 \cos (x+y)$
Since, $P_1 P_3=P_2 P_4$, we have $P_1 P_3^2=P_2 P_4^2$
Therefore, $2-2(\cos x \cos y-\sin x \sin y)$
$=2-2 \cos (x+y)$
Hence, $\cos (x+y)=\cos x \cos y-\sin x \sin y$
