Question

Suppose $A$ and $B$ are two angles such that $A,B\in (0,\pi )$ and satisfy $\sin A+\sin B=1$ and $\cos A+\cos B=0$, then the value of $12\cos 2A+4\cos 2B$ is

Answer 1

$\sin A+\sin B=1$ ......(1)

and $\cos A+\cos B=0$ ......(2)

$\sin A=1-\sin B$

Square on both sides we get

${\sin }^{2}A=1+{\sin }^{2}B-2\sin B$

From (2), we have

${\cos }^{2}A={\cos }^{2}B$ that is ${\sin }^{2}A={\sin }^{2}B$

So, $2\sin B=1$

$\sin B=\frac{1}{2}B=\frac{\pi }{6}$

$\sin A=\frac{1}{2}$

and $\cos A=-\frac{1}{2}$

$A=\frac{5\pi }{6}$

$12\cos 2A+4\cos 2B=\frac{12}{2}+\frac{4}{2}=8$