Question

Set $a,bϵ[-\pi ,\pi ]$ be such that cos (a - b) = 1 and cos (a + b) $=\frac{1}{e}$. The number of pairs of a, b satisfying the above system of equation is

• A. 0
• B. 1
• C. 2
• D. 4

Answer 1

The correct option is D

4

$\because cos(a-b)=cos0$

$\therefore a-b=2n\pi ,nϵI$

$a-b=-2\pi ,0,2\pi$

$\therefore a+b=2\pi +2a,2a,2a-2\pi$

$\therefore cos(a+b)=cos2a,cos2a,cos2a=\frac{1}{e}$

$y=cos2a=\frac{1}{e}$

Hence, number of solutions is 4.