Question

If $\sin (A+B)=1$, and $\sin (A-B)=\frac{1}{2}$. Then find $A$ and $B$.

Answer 1

$\sin (A+B)=1$

$\Rightarrow \sin (A+B)=\sin \frac{\pi }{2}$

$\Rightarrow A+B=\frac{\pi }{2}$ .........$\left(1\right)$

And $\sin (A-B)=\frac{1}{2}$

$\Rightarrow \sin (A-B)=\sin \frac{\pi }{6}$

$\Rightarrow A-B=\frac{\pi }{6}$ .........$\left(2\right)$

Adding eqns$\left(1\right)$ and $\left(2\right)$ we get

$A+B+A-B=\frac{\pi }{2}+\frac{\pi }{6}=\frac{3\pi +\pi }{6}=\frac{4\pi }{6}=\frac{2\pi }{3}$

$\therefore 2A=\frac{2\pi }{3}$ or $A=\frac{\pi }{3}$

Substitute $A=\frac{\pi }{3}$ in eqn$\left(1\right)$ we get

$A+B=\frac{\pi }{2}$

$\Rightarrow B=\frac{\pi }{2}-A=\frac{\pi }{2}-\frac{\pi }{3}=\frac{3\pi -2\pi }{6}=\frac{\pi }{6}$

Hence, $A=\frac{\pi }{3}$ and $B=\frac{\pi }{6}$