Question

Find sin(a+b) if cos(a)=3/5, sin(b)= -4/5, a is in quadrant 1 and b is in quadrant 3.

cos^-1(cos(a)=3/5)cos^-1 => a = 53.13010235. sin-1(sin(b)= -4/5) => -53.13010235

a + b = 0, and sin(0) = 0. But my answer of 0 is marked wrong. What am I missing?

Answer 1

`\sin^(-1)(\sin x)=x` only for
`-90^\circ\le x\le90^\circ`. The angle
`b` you found lies in quadrant 4, not 3. Subtract 90 degrees from the angle you found and you get an angle in quadrant 3 whose sine is still -4/5, so that

`b=\sin^(-1)\left(-\frac45\right)-90^\circ\approx-143.13^\circ`

or equivalently about 216 degrees (which does lie in quadrant 3).

###

Without looking for the exact values of
`a` and
`b`, we can make two observations:

• 
  `a` is in quadrant 1, so we know
  `\cos a` and
  `\sin a` are both positive, and
• 
  `b` is in quadrant 3, so we know
  `\cos b` and
  `\sin b` are both negative

Next, we find the values of
`\sin a` and
`\cos b` from the given values using the Pythagorean identity:

• 
  `\sin a=√(1-\cos^2a)=\frac45`
• 
  `\cos b=-√(1-\sin^2b)=-\frac35`

Then using the angle sum identity for sine, we get

`\sin(a+b)=\sin a\cos b+\cos a\sin b=\frac45\left(-\frac35\right)+\frac35\left(-\frac45\right)=-(24)/(25)\\eq0`