Question

The value of $\cos (x+y)$ is equal to

• A. $\cos x\cdot \cos y+\sin x\cdot \sin y$
• B. $\cos x\cdot \sin y+\sin x\cdot \cos y$
• C. $\cos x\cdot \cos y-\sin x\cdot \sin y$
• D. None of the above

Answer 1

Answer: (C)

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