Question

$1)$ If $\sin A=\frac{4}{5}$ and $\cos B=\frac{5}{13},$ where $0<A,B<\frac{\pi }{2},$ find the value of:

$\left(i\right)$ $\sin (A+B)$

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

$\left(ii\right)$ $\cos (A+B)$

$3)$ If $\sin A=\frac{4}{5}$ and $\cos B=\frac{5}{13},$ where $0<A,B<\frac{\pi }{2},$ find the value of:

$\left(iii\right)$ $\sin (A-B)$

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

$\left(iv\right)$ $\cos (A-B)$

Answer 1

$\left(i\right)$ Given: $\sin A=\frac{4}{5}$ and $\cos B=\frac{5}{13}$

Now, we know

$\cos A=\sqrt{1-{\sin }^{2}A}$ $[\because 0<A,B<\frac{\pi }{2}]$

$\Rightarrow \cos A=\frac{3}{5}$

Also, $\sin B=\sqrt{1-{\cos }^{2}B}=\frac{12}{13}$ $[\because 0<A,B<\frac{\pi }{2}]$

$\sin (A+B)=\sin A\cos B+\cos A\sin B$

$\Rightarrow \sin (A+B)=\frac{4}{5}\cdot \frac{5}{13}+\frac{3}{5}\cdot \frac{12}{13}$

$\therefore \sin (A+B)=\frac{56}{65}$

$\left(ii\right)$ Given: $\sin A=\frac{4}{5}$ and $\cos B=\frac{5}{13}$

Now, we know

$\cos A=\sqrt{1-{\sin }^{2}A}$ $[\because 0<A,B<\frac{\pi }{2}]$

$\Rightarrow \cos A=\frac{3}{5}$

Also, $\sin B=\sqrt{1-{\cos }^{2}B}=\frac{12}{13}$ $[\because 0<A,B<\frac{\pi }{2}]$

$\cos (A+B)=\cos A\cos B-\sin A\sin B$

$\Rightarrow \cos (A+B)=\frac{3}{5}\cdot \frac{5}{13}-\frac{4}{5}\cdot \frac{12}{13}$

$\therefore \cos (A+B)=-\frac{33}{65}$

$\left(iii\right)$ Given: $\sin A=\frac{4}{5}$ and $\cos B=\frac{5}{13}$

Now, we know

$\cos A=\sqrt{1-{\sin }^{2}A}$ $[\because 0<A,B<\frac{\pi }{2}]$

$\Rightarrow \cos A=\frac{3}{5}$

Also, $\sin B=\sqrt{1-{\cos }^{2}B}=\frac{12}{13}$ $[\because 0<A,B<\frac{\pi }{2}]$

$\sin (A-B)=\sin A\cos B-\cos A\sin B$

$\Rightarrow \sin (A-B)=\frac{4}{5}\cdot \frac{5}{13}-\frac{3}{5}\cdot \frac{12}{13}$

$\therefore \sin (A-B)=-\frac{16}{65}$

$\left(iv\right)$ Given: $\sin A=\frac{4}{5}$ and $\cos B=\frac{5}{13}$

Now, we know

$\cos A=\sqrt{1-{\sin }^{2}A}$ $[\because 0<A,B<\frac{\pi }{2}]$

$\Rightarrow \cos A=\frac{3}{5}$

Also, $\sin B=\sqrt{1-{\cos }^{2}B}=\frac{12}{13}$ $[\because 0<A,B<\frac{\pi }{2}]$

$\cos (A-B)=\cos A\cos B+\sin A\sin B$

$\Rightarrow \cos (A-B)=\frac{3}{5}\cdot \frac{5}{13}+\frac{4}{5}\cdot \frac{12}{13}$

$\therefore \cos (A-B)=\frac{63}{65}$