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

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

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Answer 1 · 2017-08-11T11:51:14

After using $u = arcsin (x)$ $u=Arcsin(x)$ or $x = sin u$ $x=sinu$ transforms,

$cos (arcsin x)$ $cos(arcsinx)$

= $cos u$ $cosu$

= $\sqrt{1 - {(sin u)}^{2}}$ $sqrt[1-(sinu)^2]$

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

Answer 2 · 2017-08-11T12:37:29

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

Explanation:

$cos ({sin}^{- 1} x)$ $cos(sin^-1x)$

Let ${sin}^{- 1} x = θ$ $sin^-1x=theta$

Then, $x = sin θ$ $x=sintheta$

Now put the value for $x$ $x$ in $cos ({sin}^{- 1} x)$ $cos(sin^-1x)$

$\Rightarrow cos ({sin}^{- 1} (sin θ))$ $=>cos(sin^-1(sintheta))$

So the equation becomes,

$\Rightarrow cos θ$ $=>costheta$

We know that ${sin}^{2} θ + {cos}^{2} θ = 1$ $sin^2theta+cos^2theta=1$

$\Rightarrow {cos}^{2} θ = 1 - {sin}^{2} θ$ $=>cos^2theta=1-sin^2theta$

$\Rightarrow cos θ = \sqrt{1 - {sin}^{2} θ}$ $=>costheta=sqrt(1-sin^2theta)$

We have already found that $x = sin θ,$ $x=sintheta,$ then $x^{2} = {sin}^{2} θ .$ $x^2=sin^2theta.$ Now put $x^{2}$ $x^2$ in the place for ${sin}^{2} θ .$ $sin^2theta.$

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

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