Question

Given that cos A 165 and sin B = 72, and that angles A and B are both in
Quadrant I, find the exact value of sin(A + B), in simplest radical form.

Answer 1

Given:

$\cos A=\dfrac{4}{\sqrt{65}}$

$\sin B=\dfrac{1}{\sqrt{2}}$

A and B both lies in I quadrant.

To find:

The value of $\sin (A+B)$.

Solution:

We know that, all trigonometric ratios are positive in first quadrant.

$sin^2\theta +\cos^2\theta =1$

Using this we get

$\sin^2A +\cos^2A =1$

$\sin^2A +\left(\dfrac{4}{\sqrt{65}}\right)^2 =1$

$\sin^2A +\dfrac{16}{65}=1$

$\sin^2A =1-\dfrac{16}{65}$

Taking square root on both sides, we get

$\sin A =\sqrt{\dfrac{65-16}{65}}$         [Only positive because A lies in I quadrant]

$\sin A =\sqrt{\dfrac{49}{65}}$

$\sin A =\dfrac{7}{\sqrt{65}}$

Similarly,

$\sin^2B +\cos^2B =1$

$\left(\dfrac{1}{\sqrt{2}}\right)^2 +\cos^2B =1$

$\dfrac{1}{2}+\cos^2B =1$

$\cos^2B =1-\dfrac{1}{2}$

Taking square root on both sides, we get

$\cos B =\sqrt{\dfrac{2-1}{2}}$         [Only positive because B lies in I quadrant]

$\cos B =\sqrt{\dfrac{1}{2}}$

$\cos B =\dfrac{1}{\sqrt{2}}$

Now,

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

$\sin (A+B)=\dfrac{7}{\sqrt{65}}\times \dfrac{1}{\sqrt{2}}+\dfrac{4}{\sqrt{65}}\times \dfrac{1}{\sqrt{2}}$

$\sin (A+B)=\dfrac{7}{\sqrt{130}}+\dfrac{4}{\sqrt{130}}$

$\sin (A+B)=\dfrac{11}{\sqrt{130}}$

Therefore, the exact value of $\sin (A+B)$ is $\dfrac{11}{\sqrt{130}}$.