Prove cos(sin^-1 x)= [itex]\sqrt{1-x}[/itex]

[nowayjose]

Homework Statement

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

Homework Equations

I would assume trigonometrical identities would be used to prove this.

 

[Infinitum]

Hello 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...

PS : your thread title is misleading

 

[nowayjose]

Thanks for the prompt reply!

PS : your thread title is misleading

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
$\sqrt{1-x^2} / 1$

 

[HallsofIvy]

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

 

[HallsofIvy]

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.

 

[rhythmiccycle]

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!

 

[nowayjose]

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:

$\sqrt{1-x^2} / 1$

 

[nowayjose]

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

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..

 

[Bohrok]

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

 

[Mentallic]

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.