If $\sin A=\dfrac{4}{5},$ $\dfrac{\pi }{2}<A<\pi$, and $\sin B=-\dfrac{2}{ \displaystyle\sqrt{5}}=-\dfrac{2\displaystyle\sqrt{5}}{5},$ $\pi <B<\dfrac{3\pi }{2}$, find the exact value of:

(a) $\cos A$

(b) $\cos B$

(c) $\cos ( A+B)$

(d) $\sin ( A+B)$

Solution  (a) We use a Pythagorean identity to find $\cos A$. $$\begin{eqnarray*} \cos A &=&\underset{\underset{\underset{\color{#0066A7}{\hbox{\(\cos A{ <0}\)}}}{\color{#0066A7}{\hbox{ A is in quadrant II}}}} {\color{#0066A7}{\displaystyle\uparrow }}}{} -\displaystyle\sqrt{1-\sin ^{2}A}=-\displaystyle\sqrt{1-\left( \dfrac{4}{5}\right) ^{2}}=-\displaystyle\sqrt{1-\dfrac{16}{25}}=-\displaystyle\sqrt{\dfrac{9}{25}}=- \dfrac{3}{5}\\ \end{eqnarray*}$$

(b) We also find $\cos B$ using a Pythagorean identity. $$\cos B=-\displaystyle\sqrt{1-\sin ^{2}B}=-\displaystyle\sqrt{1-\dfrac{4}{5}}=-\displaystyle\sqrt{\dfrac{1}{5}}=- \dfrac{\displaystyle\sqrt{5}}{5}$$

(c) We use the results from (a) and (b) and a sum formula to find $\cos ( A+B)$. $$\begin{eqnarray*} \cos ( A+B) &=&\cos A\cos B-\sin A\sin B \\[4pt] &=&\left( -\dfrac{3}{5}\right) \left( -\dfrac{\displaystyle\sqrt{5}}{5}\right) -\left( \dfrac{4}{5}\right) \left( -\dfrac{2\displaystyle\sqrt{5}}{5}\right) =\dfrac{11\displaystyle\sqrt{5}}{ 25} \end{eqnarray*}$$

(d) We use a sum formula to find $\sin (A+B)$. $$\sin ( A+B) =\sin A\cos B+\cos A\sin B=\left(\! \dfrac{4}{5}\!\right) \left(\! -\dfrac{\displaystyle\sqrt{5}}{5}\!\right) +\left(\! -\dfrac{3}{5}\!\right) \left(\! - \dfrac{2\displaystyle\sqrt{5}}{5}\!\right) =\dfrac{2\displaystyle\sqrt{5}}{25}$$