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.

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

$b=\sin^{-1}\left(-\dfrac45\right)-90^\circ\approx-143.13^\circ$

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

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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:

• $\sin a=\sqrt{1-\cos^2a}=\dfrac45$
• $\cos b=-\sqrt{1-\sin^2b}=-\dfrac35$

Then using the angle sum identity for sine, we get

$\sin(a+b)=\sin a\cos b+\cos a\sin b=\dfrac45\left(-\dfrac35\right)+\dfrac35\left(-\dfrac45\right)=-\dfrac{24}{25}\neq0$