- #1

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## Homework Statement

cos(sin

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

## Homework Equations

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

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- Thread starter nowayjose
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- #1

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cos(sin

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

- #2

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

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 :uhh:

- #3

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PS : your thread title is misleading :uhh:

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]

- #4

HallsofIvy

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

- #5

HallsofIvy

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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.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]

- #6

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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!

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!

- #7

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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]

- #8

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

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..

Last edited:

- #9

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Since sinx and sin

sin

- #10

Mentallic

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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 [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.

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