Question

If $\cos A=-\frac{24}{25}\mathrm{and}\cos B=\frac{3}{5}$, where π < A < $\frac{3\mathrm{\pi }}{2}\mathrm{and}\frac{3\mathrm{\pi }}{2}$< B < 2π, find the following:

(i) sin (A + B)
(ii) cos (A + B)

Answer 1

$$\begin{gathered} \text{Given:} \\ \cos A=-\frac{24}{25}\mathrm{and}\cos B=\frac{3}{5} \\ \text{and}\mathrm{\pi }<A<\frac{3\mathrm{\pi }}{2}\mathrm{and}\frac{3\mathrm{\pi }}{2}<B<2\mathrm{\pi }. \\ \text{That is,}A\mathrm{is}\mathrm{in}\mathrm{third}\mathrm{quadrant}\mathrm{and}\mathrm{B}\mathrm{is}\mathrm{in}\mathrm{fourth}\mathrm{qudrant}. \\ \mathrm{We}\mathrm{know}\mathrm{that}\sin e\mathrm{function}\mathrm{is}\mathrm{negative}\mathrm{in}\mathrm{third}\mathrm{and}\mathrm{fourth}\mathrm{quadrant}s. \\ \mathrm{Therefore}, \\ \sin A=-\sqrt{1-{\cos }^{2}A}\mathrm{and}\sin B=-\sqrt{1-{\cos }^{2}B} \\ \Rightarrow \sin A=\sqrt{1-{\left(\frac{-24}{25}\right)}^2}\mathrm{and}\sin B=-\sqrt{1-{\left(\frac{3}{5}\right)}^2} \\ \Rightarrow \sin A=-\sqrt{1-\frac{576}{625}}\mathrm{and}\sin B=-\sqrt{1-\frac{9}{25}} \\ \Rightarrow \sin A=-\sqrt{\frac{49}{625}}\mathrm{and}\sin B=-\sqrt{\frac{16}{25}} \\ \Rightarrow \sin A=\frac{-7}{25}\mathrm{and}\sin B=\frac{-4}{5} \end{gathered}$$

$$\begin{gathered} \mathrm{Now}, \\ \left(\mathrm{i}\right)\sin \left(A+B\right)=\sin A\cos B+\cos A\sin B \\ =\frac{-7}{25}\times \frac{3}{5}+\frac{-24}{25}\times \frac{-4}{5} \\ =\frac{-21}{125}+\frac{96}{125} \\ =\frac{75}{125} \end{gathered}$$
$=\frac{3}{5}$

$$\begin{gathered} \left(\mathrm{ii}\right)\cos \left(A+B\right)=\cos A\cos B-\sin A\sin B \\ =\frac{-24}{25}\times \frac{3}{5}-\frac{-7}{25}\times \frac{-4}{5} \\ =\frac{-72}{125}-\frac{28}{125} \\ =\frac{-100}{125} \\ =\frac{-4}{5} \end{gathered}$$