1. ## cos((sinx)^-1)

i was wondering if anyone could help me to simplify cos((sinx)^-1) i remember someone saying in class that it was something like
2sqareroot(9-x^2) but i don't have a clue how to get there. any help would be much apreciated

2. Let $\theta = sin^{-1}(x)$. You can think of this as the angle whose sine is x/1. In a right triangle, sine is opposite over hypotenuse, so draw a right triangle, label one of the acute angles as theta, the opposite side as x and the hypotenuse as 1. Use the Pythagorean theorem to find the missing (adjacent) side:
$a^{2} + b^{2} = c^{2}$
$x^{2} + b^{2} = 1^{2}$
Solve for b. Label this on the adjacent side.

Now $cos(\theta) = cos(sin^{-1}(x))$, and cosine is adjacent over hypotenuse. Can you find $cos(sin^{-1}(x))$ now?

01

3. Or... :

$\cos(t)=\pm \sqrt{1-\sin^2(t)}$ (from the well-known identity $\cos^2(t)+\sin^2(t)=1$

But since the arcsin ranges from $-\pi/2$ to $\pi/2$, $\cos(\arcsin(x))\geq 0$

And hence $\cos(\arcsin(x))=\sqrt{1-\sin^2(\arcsin(x))}=\sqrt{1-x^2}$