How do you simplify $cos ({sin}^{- 1} x)$ $cos(sin^-1 x)$ ?

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How do you simplify $cos ({sin}^{- 1} x)$ $cos(sin^-1 x)$ ?

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Answer 1 · 2016-10-21T11:12:10

± $\sqrt{1 - x^{2}}$ $sqrt (1-x^2)$

Explanation:

$cos ({sin}^{- 1} x)$ $cos(sin^-1 x)$
Let,
${sin}^{- 1} x = θ$ $sin^-1x = theta$
$\Rightarrow sin θ = x$ $=>sin theta = x$
$\Rightarrow {sin}^{2} θ = x^{2}$ $=>sin^2theta =x^2$
$\Rightarrow 1 - {cos}^{2} θ = x^{2}$ $=>1-cos^2theta = x^2$
$\Rightarrow {cos}^{2} θ = 1 - x^{2}$ $=>cos^2theta = 1-x^2$
$\Rightarrow cos θ = \pm \sqrt{1 - x^{2}}$ $=>cos theta =± sqrt (1-x^2)$
$\Rightarrow θ = {cos}^{- 1} \pm \sqrt{1 - x^{2}}$ $=>theta =cos^-1±sqrt(1-x^2)$
Putting this,
$cos ({cos}^{- 1} \pm \sqrt{1 - x^{2}})$ $cos(cos^-1±sqrt(1-x^2))$
$= \pm \sqrt{1 - x^{2}}$ $=±sqrt(1-x^2)$

But ${sin}^{- 1} x$ $sin^(-1)x$ is, by definition, in $[- \frac{π}{2}, \frac{π}{2}]$ $[-pi/2,pi/2]$ so $cos ({sin}^{- 1} x) \geq 0$ $cos(sin^-1x) >= 0$

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

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How do you simplify $cos ({sin}^{- 1} x)$ $cos(sin^-1 x)$ ?

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