`cos(sin^-1(x)) = sqrt(1 - x^2)` Verfiy the identity.

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Textbook Question

Chapter 5, 5.2 - Problem 48 - Precalculus (3rd Edition, Ron Larson).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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For both sides to have sense, x must be in [-1, 1].

For such x'es, `sin^(-1)(x)` is the angle y in `[-pi/2, pi/2]` such that `sin(y) = x.`

But `cos^2(y) + sin^2(y) = 1,` so `cos^2(y) = 1 - sin^2(y) = 1-x^2.`

Also, for y in `[-pi/2, pi/2]` always cos(y)>=0. Therefore we can state that

`cos(y) = cos(sin^(-1)(x)) = sqrt(1-x^2)`  (not `-sqrt(1-x^2)` ), QED.

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balajia | College Teacher | (Level 1) eNoter

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Taking x as opposite sid and 1 as hypotenuse we get adjacent side as `sqrt(1-x^2).`

`cos(sin^-1x) =cos(cos^-1(sqrt(1-x^2)/1)) = sqrt(1-x^2).`

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