Question

If $\sin A=\frac{4}{5},\frac{\pi }{2}<A<\pi$ and $\cos B=\frac{5}{13},\frac{3\pi }{2}<B<2\pi$, find
(i) $\sin (A+B)$, (ii)$\cos (A-B)$, (iii)$\tan (A-B)$

Answer 1

$\sin A=\frac{4}{5},\cos B=\frac{5}{13}$

$\sin A=\frac{4}{5},Aϵ(\pi /2,\pi )$

$⟹\cos A=\frac{3}{\sqrt{25-16}}=\frac{-3}{5}$

$\cos B=\frac{5}{13}$$,Bϵ(3\pi /2,2\pi )$

So, $\sin B=\frac{-12}{13}$

$\left(1\right)$ $\sin (A+B)=\sin A\cos B+\cos A\sin B=\left(\frac{4}{5}\right)\left(\frac{5}{13}\right)+\left(\frac{-3}{5}\right)\left(\frac{-12}{13}\right)=\frac{56}{65}$

$\left(2\right)$ $\cos (A-B)=\cos A\cos B+\sin A\sin B=\left(\frac{-3}{5}\right)\left(\frac{5}{13}\right)+\left(\frac{4}{5}\right)\left(\frac{-12}{13}\right)=\frac{-63}{65}$

$\left(3\right)$ $\sin (A-B)=\sin A\cos B-\cos A\sin B=\left(\frac{4}{5}\right)\left(\frac{5}{13}\right)-\left(\frac{-3}{5}\right)\left(\frac{-12}{13}\right)=\frac{-16}{65}$

$\left(4\right)$ $\tan (A-B)=\frac{\sin A-B}{\cos (A-B)}=\frac{(-16/65)}{(-63/65)}=\frac{16}{63}$