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=(4)/(√(65))`

`\sin B=(1)/(√(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((4)/(√(65))\right)^2 =1`

`\sin^2A +(16)/(65)=1`

`\sin^2A =1-(16)/(65)`

Taking square root on both sides, we get

`\sin A =\sqrt{(65-16)/(65)}` [Only positive because A lies in I quadrant]

`\sin A =\sqrt{(49)/(65)}`

`\sin A =(7)/(√(65))`

Similarly,

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

`\left((1)/(√(2))\right)^2 +\cos^2B =1`

`(1)/(2)+\cos^2B =1`

`\cos^2B =1-(1)/(2)`

Taking square root on both sides, we get

`\cos B =\sqrt{(2-1)/(2)}` [Only positive because B lies in I quadrant]

`\cos B =\sqrt{(1)/(2)}`

`\cos B =(1)/(√(2))`

Now,

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

`\sin (A+B)=(7)/(√(65))* (1)/(√(2))+(4)/(√(65))* (1)/(√(2))`

`\sin (A+B)=(7)/(√(130))+(4)/(√(130))`

`\sin (A+B)=(11)/(√(130))`

Therefore, the exact value of
`\sin (A+B)` is
`(11)/(√(130))`.