Question

(a) If $\sin A=\frac{12}{13}\mathrm{and}\sin B=\frac{4}{5}$, where $\frac{\mathrm{\pi }}{2}$< A < π and 0 < B < $\frac{\mathrm{\pi }}{2}$, find the following:

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

(b) If $\sin A=\frac{3}{5},\cos B=-\frac{12}{13}$, where A and B both lie in second quadrant, find the value of sin (A + B).

Answer 1

$$\begin{gathered} \left(\mathrm{a}\right)\mathrm{Given}: \\ \sin A=\frac{12}{13}\mathrm{and}\sin B=\frac{4}{5} \\ \mathrm{When},\frac{\mathrm{\pi }}{2}<A<\mathrm{\pi }\mathrm{and}0<B<\frac{\mathrm{\pi }}{2}, \\ \cos A=-\sqrt{1-{\sin }^{2}A}\mathrm{and}\cos B=\sqrt{1-{\sin }^{2}B} \\ \left(\mathrm{As}\cos \text{ine}\mathrm{function}\mathrm{is}\mathrm{negative}\mathrm{in}\mathrm{second}\mathrm{qudrant}\mathrm{and}\mathrm{positive}\mathrm{in}\mathrm{first}\mathrm{quadrant}\right) \end{gathered}$$

$$\begin{gathered} \Rightarrow \cos A=-\sqrt{1-{\left(\frac{12}{13}\right)}^2}\mathrm{and}\cos B=\sqrt{1-{\left(\frac{4}{5}\right)}^2} \\ \Rightarrow \cos A=-\sqrt{1-\frac{144}{169}}\mathrm{and}\cos B=\sqrt{1-\frac{16}{25}} \\ \Rightarrow \cos A=-\sqrt{\frac{25}{169}}\mathrm{and}\cos B=\sqrt{\frac{9}{25}} \\ \Rightarrow \cos A=\frac{-5}{13}\mathrm{and}\cos B=\frac{3}{5} \\ \mathrm{Now}, \end{gathered}$$


$$\begin{gathered} \left(\mathrm{i}\right)\sin \left(A+B\right)=\sin A\cos B+\cos A\sin B \\ =\frac{12}{13}\times \frac{3}{5}+\frac{-5}{13}\times \frac{4}{5} \\ =\frac{36}{65}+\frac{-20}{65} \\ =\frac{16}{65} \\ \left(\mathrm{ii}\right)\cos \left(A+B\right)=\cos A\cos B-\sin A\sin B \\ =\frac{-5}{13}\times \frac{3}{5}-\frac{12}{13}\times \frac{4}{5} \\ =\frac{-15}{65}-\frac{48}{65} \\ =\frac{-63}{65} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{b}\right)\text{Given:} \\ \sin A=\frac{3}{5}and\cos B=-\frac{12}{13} \\ \text{and}\mathrm{that}A\mathrm{and}B\mathrm{both}\mathrm{lie}\mathrm{in}\mathrm{second}\mathrm{qudrant}. \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{in}\mathrm{second}\mathrm{quadrant}\sin e\mathrm{function}\mathrm{is}\mathrm{positive}\mathrm{and}\cos \text{ine}\mathrm{function}\mathrm{is}\mathrm{negative}. \\ \mathrm{Therefore}, \\ \cos A=-\sqrt{1-{\sin }^{2}A}\mathrm{and}\sin B=\sqrt{1-{\cos }^{2}B} \\ \Rightarrow \cos A=-\sqrt{1-{\left(\frac{3}{5}\right)}^2}\mathrm{and}\sin B=\sqrt{1-{\left(\frac{-12}{13}\right)}^2} \\ \Rightarrow \cos A=-\sqrt{1-\frac{9}{25}}\mathrm{and}\sin B=\sqrt{1-\frac{144}{169}} \\ \Rightarrow \cos A=-\sqrt{\frac{16}{25}}\mathrm{and}\sin B=\sqrt{\frac{25}{69}} \\ \Rightarrow \cos A=\frac{-4}{5}\mathrm{and}\sin B=\frac{5}{13} \\ \mathrm{Now}, \\ \sin \left(A+B\right)=\sin A\cos B+\cos A\sin B \\ =\frac{3}{5}\times \frac{-12}{13}+\frac{-4}{5}\times \frac{5}{13} \\ =\frac{-36}{65}-\frac{20}{65} \\ =\frac{-56}{65} \end{gathered}$$