Question

If $\cos A=\frac{4}{5},\cos B=\frac{12}{13},\frac{3\pi }{2}<A,B<2\pi$, find the values of the following.
(i) $\cos (A+B)$
(ii) $\sin (A-B)$

Answer 1

since A and B both lie in the IV quadrant, it follows that $\sin A$ and $\sin B$ are negative. Therefore,

$\sin A=-\sqrt{1-{\cos }^{2}A}$

$\Rightarrow \sin A=-\sqrt{1-\frac{16}{25}}=-\frac{3}{5}$

and $\sin B=-\sqrt{1-{\cos }^{2}B}$

$\Rightarrow \sin B=-\sqrt{1-\frac{144}{169}}=-\frac{5}{13}$

Now,

(i) $\cos (A+B)=\cos A\cos B-\sin A\sin B$

$=\frac{4}{5}\times \frac{12}{13}-(\frac{-4}{13}\times \frac{-5}{13})=\frac{33}{65}$

(ii) $\sin (A-B)=\sin A\cos B-\cos A\sin B$

$=\frac{-3}{5}\times \frac{12}{13}-\frac{4}{5}\times \frac{-5}{13}=\frac{-16}{25}$