Question

How to geometrically prove that $\cos (x+y)=\cos x\cos y-\sin x\sin y$

Answer 1

To prove :$\cos (x+y)=\cos x\cos y-\sin x\sin y$
Consider the unit circle with centre at origin.
Let x be the angle $P_{4}O{P}_1$ and y be angle $P_{1}O{P}_2$ then $(x+y)$ is angle $P_{4}O{P}_2$.
Let (-y)be angle $P_{4}O{P}_3$ then $P_1,P_2,P_{3}and{P}_4$ woill have coordinates
$P_1$ $(\cos x,\sin x)$
$P_2$ $\left(cos\right(x+y),sin(x+y\left)\right)$
$P_3$ $\left(cos\right(-y),sin(-y\left)\right)$
$P_4$ $(1,0)$

Consider triangles $P_{1}O{P}_3$ and $P_{2}O{P}_4$ (Both are congruent)
$\therefore P_1{P}_3=P_2{P}_4$
Using distance formula
$P_1{P}_3^2=[\cos x-\cos (-y){]}^2+[\sin x-\sin (-y){]}^2$
$=[\cos x-\cos (y){]}^2+[\sin x+\sin \left(y\right){]}^2$
$={\cos }^{2}x+{\cos }^{2}y-2\cos x\cos y+{\sin }^{2}x+{\sin }^{2}y+2\sin x\sin y$
[$\because {\sin }^{2}x+{\cos }^{2}y=1$]
$=2-2(\cos x\cos y-\sin x\sin y)$
Similarly,
$P_2{P}_4^2=\left(\cos \right(x+y)-1{)}^2+{\sin }^2(x+y)$
$={\cos }^2(x+y)+1-2\cos (x+y)+{\sin }^2(x+y)$
$=2-2\cos (x+y)$
$P_2{P}_4^2=2[1-\cos (x+y\left)\right]$
$\because P_1{P}_3=P_2{P}_4$
$\therefore P_1{P}_3^2=P_2{P}_4^2$
$\therefore 2-2(\cos x\cos y-\sin x\sin y)=2-2\cos (x+y)$
$\Rightarrow 2(\cos x\cos y-\sin x\sin y)=2\cos (x+y)$
$\therefore \cos (x+y)=\cos x\cos y-\sin x\sin y$

Hence Proved.