Question

If $\sin {\theta }_1-\sin {\theta }_2=a$ and $\cos {\theta }_1-\cos {\theta }_2=b$, then

• A. $a^2+b^2\ge 4$
• B. $a^2+b^2\le 4$
• C. $a^2+b^2\ge 3$
• D. $a^2+b^2\le 2$

Answer 1

The correct option is B $a^2+b^2\le 4$

$\sin {\theta }_1-\sin {\theta }_2=a⟶\left(1\right)$

$\cos {\theta }_1-\cos {\theta }_2=b⟶\left(2\right)$

Squaring and adding (1) and (2)
$a^2+b^2=2(1-\cos ({\theta }_1-{\theta }_2\left)\right)$

and, $-1<\cos ({\theta }_1-{\theta }_2)<1$

Therefore,
$0\le a^2+b^2\le 4$

Hence, option 'C' is correct.