Question

Find Sin (A+B) When SinA=(3/5) with 90 degrees < A < 270 degrees, and if CosB=(8/17) with 0 degrees < B< 180 degrees

Answer 1

Recall the compound angle identity:

`\sin(a+b)=\sin a\cos b+\cos a\sin b`

`90^\circ<a<270^\circ`, so we expect
`\cos a<0`, and
`0^\circ<b<180^\circ`, so we expect
`\sin b>0`. Then by the Pythagorean identity,

`\cos^2x+\sin^2x=1\implies\begin{cases}\cos a=-√(1-\sin^2a)=-\frac45\\\\\sin b=√(1-\cos^2b)=(15)/(17)\end{cases}`

Then

`\sin(a+b)=-(36)/(85)`