$\cos x = 0,$

$x = \frac{\pi}{2}, \frac{3\pi}{2}$

$\cos(x) = \cos\ (x+k. 2\pi)$

$x = \biggl ( n+ \frac{1}{2}\biggl) {\pi}$ ,n is integer.

$\cos {\theta} = \frac {AB}{AC}$

$\sin \biggl (\frac {\pi}{2}-{\theta} \biggl ) = \frac {AB}{AC}$

$\cos {\theta} = \sin \biggl (\frac {\pi}{2}-{\theta} \biggl )$

$\cos \biggl (\frac {\pi}{4}\biggl ) = \frac {1} {\sqrt{2} }$

$\sin \biggl (\frac {\pi}{4} \biggl ) = \frac {1}{\sqrt{2} }$

$\triangle$ ABC and $\triangle$ ABD are congruent

$\frac {\pi}{3} + {\theta}+{\theta}= {\pi},\ {\theta}= \frac {\pi} {3}$

$\Rightarrow \triangle ADC$ is equilateral

$\Rightarrow CD= 1$ unit

$\sin \biggl (\frac {\pi}{6} \biggl )= \frac {1}{2}$

$\sin (x) = b$

$\sin (-x) = d$

$\triangle OAP \cong \triangle OAQ$

$c = a$

$d = -b$

$\sin (-x) = -b = -\sin (x)$

$\sin (-x) = -\sin(x)$

$f (-x) = -f(x)$ $\Leftrightarrow$ Odd functions

$\cos (x) = a$

$\cos (-x) = a$

$\cos (x) = \cos (-x)$

$f (x) = f (-x)$ $\Leftrightarrow$ Even functions

$\triangle OPQ \cong \triangle OAR$

$OQ=OR$

$OP=OA$

$\angle POQ= \angle AOR$

$\angle POQ= x-y$

$QP ^2= AR ^2$

$QP ^2= (\cos x - \cos y) ^2 + (\sin x - \sin y) ^2$

$AR ^2= [\cos (x -y) - 1] ^2 + \sin ^2 (x-y)$

As,$QP ^2 = (\cos x - \cos y) ^2 + (\sin x - \sin y ) ^2$

$= \cos ^2 x + \cos ^2 y -2 \cos x \cos y + \sin ^2 x + \sin ^2 y -2 \sin x \sin y$

$= 1 + 1 -2 \cos x \cos y -2 \sin x \sin y$

$AR ^2 = [\cos (x- y) -1] ^2 + \sin ^2 (x -y)$

$= \cos ^2 (x- y) +1 -2 \cos (x -y) + \sin ^2 (x-y)$

$= 1 + 1 -2 \cos (x -y )$

$\because QP ^2 = AR ^2$

$\Rightarrow \cos (x- y) = \cos x \cos y + \sin x \sin y$

$\because \cos (x- y) = \cos x \cos y + \sin x \sin y$

$\cos (x+ y) = \cos (x - (- y))$

$\cos (x+ y) = \cos x \cos (- y) + \sin (x) \sin (-y)$

$\cos (x+ y) = \cos x \cos y - \sin (x) \sin (y)$

Thank you