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

nowayjose

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

cos(sin^-1x) = [itex]\sqrt{1-x^2}[/itex]

2. Relevant equations

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

Infinitum

Hello nowayjose!

Yes, they would...

Why don't you start by assuming [itex]\theta = sin^{-1}x[/itex], 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!
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...).

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

HallsofIvy

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

HallsofIvy

What you have written here makes little sense. If [itex]\theta= sin^{-1}(x)[/itex] then, yes, [itex]sin(\theta)= x[/itex], 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!

Attached Thumbnails

 

nowayjose

People sorry for the typo and for not being clear about my thought process.

Bascially what i meant was:

[itex]\theta = sin^{-1}x[/itex]

[itex]sin\theta = x[/itex]

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 [itex]\sqrt{1-x^2}[/itex]

The cosine angle is the quotient of the adjacent and the hypotenuse:

[itex]\sqrt{1-x^2} / 1[/itex]

nowayjose

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

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 [itex]y=\sin(x)[/itex] and you want to make a process to get back to just x, and for this case its inverse will be [itex]\sin^{-1}(y)[/itex].
Other inverses are, for example, the inverse of [itex]y=x^2[/itex] is [itex]\sqrt{y}[/itex] because [itex]\sqrt{x^2}=x[/itex] (technically it's |x| so that's why we specify domains, in this case [itex]x\geq 0[/itex])
Another would include [itex]y=\ln(x)[/itex] and [itex]e^y[/itex]

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

You multiply the indices, not add.