Thread: Analytic Trig question 1

Find the exact value.

18) cos[sin^-1(-(sqrt3)/2)]

sqrt(3)/2 is a famous trigonometical value of some angle.

cos^2(x) + sin^2(x) = 1 --> you can express cos with sin.
And sin(sin^-1) is identic function, isn't it?

so the angle value is 30 degrees. Does that mean the answer is [(sqrt3)/2], [1/2]?

How have you gotten the first answer? The second is correct.

The answer was derived from the graph using 30 degrees. What is the answer?

Originally Posted by kukid123

Find the exact value.

18) cos[sin^-1(-(sqrt3)/2)]

Hi kukid123,

Here is one way of doing it:

$\displaystyle \cos(\sin^{-1}(-\frac{\sqrt{3}}{2}))$

$\displaystyle \sin^{-1}(-\frac{\sqrt{3}}{2})=180^0+60^0=240^0$ 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 $\displaystyle 0<\theta<360^0$.
So $\displaystyle \cos(\sin^{-1}(-\frac{\sqrt{3}}{2}))=\cos 240^0 = -\frac{1}{2}$ or $\displaystyle \cos(\sin^{-1}(-\frac{\sqrt{3}}{2}))=\cos 300^0 = \frac{1}{2}$