Using Trigonometry Identities

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 \(\sin (A+B) \). \[ \sin ( A+B) =\sin A\cos B+\cos A\sin B=\left(\! \dfrac{4}{5}\!\right) \left(\! -\dfrac{\displaystyle\sqrt{5}}{5}\!\right) +\left(\! -\dfrac{3}{5}\!\right) \left(\! - \dfrac{2\displaystyle\sqrt{5}}{5}\!\right) =\dfrac{2\displaystyle\sqrt{5}}{25} \]