prove cos(sin^-1 x)= $\sqrt{1-x}$

1. The problem statement, all variables and given/known data

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

2. Relevant equations

I would assume trigonometrical identities would be used to prove this.
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Hello nowayjose!

 Quote by nowayjose I would assume trigonometrical identities would be used to prove this.
Yes, they would...

Why don't you start by assuming $\theta = sin^{-1}x$, and then draw out a triangle to find a relation between theta and cosine, that you can use...

Sorry, and the question's undoubtedly stupid. I've used this method before and haven't happened to used any identities (or so i believe...).

$\theta = sin^{-1}x$
$sin\theta = x$
$sin = 1/X$
the cosine side must therefore be $\sqrt{1-x^2}$
therefore the cosine angle is
$\sqrt{1-x^2} / 1$

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prove cos(sin^-1 x)= $\sqrt{1-x}$

Or, different wording of the same idea: $sin^2(\theta)+ cos^2(\theta)= 1$ so that $cos(\theta)= \pm\sqrt{1- sin^2(\theta)}$. So
$$cos(sin^{-1}(x))= \pm\sqrt{1- sin^2(sin^{-1}(x)}= \pm\sqrt{1- x^2}$$

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 Quote by nowayjose Thanks for the prompt reply! Sorry, and the question's undoubtedly stupid. I've used this method before and haven't happened to used any identities (or so i believe...). $\theta = sin^{-1}x$ $sin\theta = x$ $sin = 1/X$ the cosine side must therefore be $\sqrt{1-x^2}$ therefore the cosine angle is $\sqrt{1-x^2} / 1$
What you have written here makes little sense. If $\theta= sin^{-1}(x)$ then, yes, $sin(\theta)= x$, but you cannot write "sin" without some argument. And the "-1" does NOT indicate reciprocal (1/x), it means the inverse function.
 consider the attached triangle picture (sorry its sloppy) in that case Sin(theta) = x (hypotenuse is 1, opposite is x) thus sin^-1(x) = theta. For that same theta, using a^2 + b^2 = c^2...... x^2 + b^2 = 1^2 b^2 = 1-x^2 b = sqrt( 1 - x^2) and cos(theta) = adj / hyp so, cos(theta) = sqrt( 1 - x^2) / 1 cos(theta) = sqrt( 1 - x^2) recall: sin^-1(x) = theta so sub in theta cos(sin^-1(x)) = sqrt( 1 - x^2) Proved! Attached Thumbnails
 People sorry for the typo and for not being clear about my thought process. Bascially what i meant was: $\theta = sin^{-1}x$ $sin\theta = x$ If the sine angle is X, then the opposite is X and the hypotenuse 1. the adjacent side can now be calculated using pythagoras, which gives $\sqrt{1-x^2}$ The cosine angle is the quotient of the adjacent and the hypotenuse: $\sqrt{1-x^2} / 1$

 Quote by HallsofIvy $$cos(sin^{-1}(x))= \pm\sqrt{1- sin^2(sin^{-1}(x)}= \pm\sqrt{1- x^2}$$
Would you care explaining me how you multiplied out what's rooted?

I know sin(sin^-1) cancel out because you add the indices, so shouldn't that leave sin x..
 Those are inverse functions, so they don't simplify like exponents do. Since sinx and sin-1x are inverses and sin2x = (sinx)2, sin(sin-1x) = x and sin2(sin-1x) = (sin(sin-1x))2 = x2

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 Quote by nowayjose I know sin(sin^-1) cancel out because you add the indices, so shouldn't that leave sin x..
No, don't think of it that way. Inverse functions are a fancy word of saying "doing the opposite". You have some function, such as $y=\sin(x)$ and you want to make a process to get back to just x, and for this case its inverse will be $\sin^{-1}(y)$.
Other inverses are, for example, the inverse of $y=x^2$ is $\sqrt{y}$ because $\sqrt{x^2}=x$ (technically it's |x| so that's why we specify domains, in this case $x\geq 0$)
Another would include $y=\ln(x)$ and $e^y$

Also, keep in mind that if you have a function $y=x^n$ and applying its inverse $$\sqrt[n]{y}=y^{1/n}$$ the reason we get back to x is because $$\left(x^n\right)^{1/n}=x^{n\cdot\frac{1}{n}}=x^{\frac{n}{n}}=x$$

You multiply the indices, not add.
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