Analytic Trig question 1

**kukid123** · Dec 15th 2008, 10:05 AM

Find the exact value.

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

**Skalkaz** · Dec 15th 2008, 10:13 AM

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?

**kukid123** · Dec 15th 2008, 10:23 AM

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

**Skalkaz** · Dec 15th 2008, 10:33 AM

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

**kukid123** · Dec 16th 2008, 09:14 AM

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

**chabmgph** · Dec 16th 2008, 10:11 AM

Originally Posted by kukid123

Find the exact value.

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

Hi kukid123,

Here is one way of doing it:

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

$\sin^{-1}(-\frac{\sqrt{3}}{2})=180^0+60^0=240^0$ or $\sin^{-1}(-\frac{\sqrt{3}}{2})=360^0-60^0=300^0$ Since sine is negative in the 3rd and 4th quadrants if $0<\theta<360^0$ .
So $\cos(\sin^{-1}(-\frac{\sqrt{3}}{2}))=\cos 240^0 = -\frac{1}{2}$ or $\cos(\sin^{-1}(-\frac{\sqrt{3}}{2}))=\cos 300^0 = \frac{1}{2}$

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