Question

If $\sin \mathrm{A}=\frac{4}{5}$and $\cos \mathrm{B}=\frac{5}{13}$, where 0 < A, $\mathrm{B}<\frac{\mathrm{\pi }}{2}$, find the values of the following:

(i) sin (A + B)
(ii) cos (A + B)
(iii) sin (A − B)
(iv) cos (A − B)

Answer 1

$$\begin{gathered} \text{Given}: \\ \sin A=\frac{4}{5}\mathrm{and}\cos B=\frac{5}{13} \\ \mathrm{We}\mathrm{know}\mathrm{that} \\ \cos A=\sqrt{1-{\sin }^{2}A}\mathrm{and}\sin B=\sqrt{1-{\cos }^{2}B},\mathrm{where}0<A,B<\frac{\mathrm{\pi }}{2} \\ \Rightarrow \cos A=\sqrt{1-{\left(\frac{4}{5}\right)}^2}\mathrm{and}\sin B=\sqrt{1-{\left(\frac{5}{13}\right)}^2} \\ \Rightarrow \cos A=\sqrt{1-\frac{16}{25}}\mathrm{and}\sin B=\sqrt{1-\frac{25}{169}} \\ \Rightarrow \cos A=\sqrt{\frac{9}{25}}\mathrm{and}\sin B=\sqrt{\frac{144}{169}} \\ \Rightarrow \cos A=\frac{3}{5}\mathrm{and}\sin B=\frac{12}{13} \end{gathered}$$

Now,

$$\begin{gathered} \left(\mathrm{i}\right)\sin \left(A+B\right)=\sin A\cos B+\cos A\sin B \\ =\frac{4}{5}\times \frac{5}{13}+\frac{3}{5}\times \frac{12}{13} \\ =\frac{20}{65}+\frac{36}{65} \\ =\frac{56}{65} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\cos \left(A+B\right)=\cos A\cos B-\sin A\sin B \\ =\frac{3}{5}\times \frac{5}{13}-\frac{4}{5}\times \frac{12}{13} \\ =\frac{15}{65}-\frac{48}{55} \\ =\frac{-33}{65} \end{gathered}$$


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


$$\begin{gathered} \left(\text{iii}\right)\sin \left(A-B\right)=\sin A\cos B-\cos A\sin B \\ =\frac{4}{5}\times \frac{5}{13}-\frac{3}{5}\times \frac{12}{13} \\ =\frac{20}{65}-\frac{36}{65} \\ =\frac{-16}{65} \end{gathered}$$

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