identities (trig)

**dee18** · September 30th 2008, 09:18 AM

prove that: cos (sin^(-1) x) = (1-x^2)^(1/2)
please help. I don't even know where to start.

**Chop Suey** · September 30th 2008, 09:54 AM

$\sin^{-1}{x}$ is an angle. Therefore:

$\sin^{-1}{x} = \theta$

$\Rightarrow x = \sin{\theta}$

Recall that in right triangle trigonometry:
$\sin{\theta} = \frac{\text{opp}}{\text{hyp}}$

You can find the adj side using the Pythagorean theorem. Now we can find $\cos{\theta}$ , which is:

$\cos{\theta} = \frac{\text{adj}}{\text{hyp}}$

**Chop Suey** · September 30th 2008, 09:58 AM

Originally Posted by dee18

prove that: cos (sin^(-1) x) = (1-x^2)^(1/2)
please help. I don't even know where to start.

Alternatively, if $y = \cos{(\sin^{-1}{x})}$ , then:

$\Rightarrow y^2 = \cos^2{(\sin^{-1}{x})}$

$\Rightarrow y^2 = 1 - \sin^2{(\sin^{-1}{x})}$

$\Rightarrow y = \ldots$

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