# If sin

Question:

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

(i) sin (A + B)

(ii) cos (A + B)

(iii) sin (A − B)

(iv) cos (A − B)

Solution:

Given :

$\sin A=\frac{4}{5}$ and $\cos B=\frac{5}{13}$

We know that

$\cos A=\sqrt{1-\sin ^{2} A} \quad$ and $\quad \sin B=\sqrt{1-\cos ^{2} B} \quad, \quad$ where $0$\Rightarrow \cos A=\sqrt{1-\left(\frac{4}{5}\right)^{2}} \quad$and$\quad \sin B=\sqrt{1-\left(\frac{5}{13}\right)^{2}}\Rightarrow \cos A=\sqrt{1-\frac{16}{25}} \quad$and$\quad \sin B=\sqrt{1-\frac{25}{169}}\Rightarrow \cos A=\sqrt{\frac{9}{25}} \quad$and$\sin B=\sqrt{\frac{144}{169}}\Rightarrow \quad \cos A=\frac{3}{5} \quad$and$\quad \sin B=\frac{12}{13}$Now, (i)$\sin (A+B)=\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}$(ii)$\cos (A+B)=\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}$(iii)$\sin (A-B)=\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}$(iv)$\cos (A-B)=\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}\$

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