Find the exact value.
18) cos[sin^-1(-(sqrt3)/2)]
Hi kukid123,
Here is one way of doing it:
$\displaystyle \sin^{-1}(-\frac{\sqrt{3}}{2})=180^0+60^0=240^0$ or $\displaystyle \sin^{-1}(-\frac{\sqrt{3}}{2})=360^0-60^0=300^0$ Since sine is negative in the 3rd and 4th quadrants if $\displaystyle 0<\theta<360^0$.$\displaystyle \cos(\sin^{-1}(-\frac{\sqrt{3}}{2}))$
So $\displaystyle \cos(\sin^{-1}(-\frac{\sqrt{3}}{2}))=\cos 240^0 = -\frac{1}{2}$ or $\displaystyle \cos(\sin^{-1}(-\frac{\sqrt{3}}{2}))=\cos 300^0 = \frac{1}{2}$