How do you simplify $cos (2 {sin}^{- 1} x)$ $cos ( 2 Sin^-1 x)$ ?

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How do you simplify $cos (2 {sin}^{- 1} x)$ $cos ( 2 Sin^-1 x)$ ?

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Answer 1 · 2016-07-07T07:03:56

$cos (2 {sin}^{- 1} x) = 1 - 2 x^{2} .$ $cos(2sin^-1x)=1-2x^2.$

Explanation:

Let ${sin}^{- 1} x = θ$ $sin^-1x=theta$ , so that, $x = sin θ, | x | \leq 1, θ \in [- \frac{π}{2}, \frac{π}{2}] .$ $x=sintheta, |x|<=1, theta in [-pi/2,pi/2].$

Now, Reqd. Value $= cos (2 {sin}^{- 1} x) = cos 2 θ = 1 - 2 {sin}^{2} θ = 1 - 2 x^{2} .$ $= cos(2sin^-1x)=cos2theta=1-2sin^2theta=1-2x^2.$

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How do you simplify $cos (2 {sin}^{- 1} x)$ $cos ( 2 Sin^-1 x)$ ?

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