How do you simplify cos(sin^-1 x)cos(sin−1x)?

1 Answer
Subha · Jim H
Oct 21, 2016

±sqrt (1-x^2)√1−x2

Explanation:

cos(sin^-1 x)cos(sin−1x)
Let,
sin^-1x = thetasin−1x=θ
=>sin theta = x⇒sinθ=x
=>sin^2theta =x^2⇒sin2θ=x2
=>1-cos^2theta = x^2⇒1−cos2θ=x2
=>cos^2theta = 1-x^2⇒cos2θ=1−x2
=>cos theta =± sqrt (1-x^2) ⇒cosθ=±√1−x2
=>theta =cos^-1±sqrt(1-x^2)⇒θ=cos−1±√1−x2
Putting this,
cos(cos^-1±sqrt(1-x^2))cos(cos−1±√1−x2)
=±sqrt(1-x^2)=±√1−x2

But sin^(-1)xsin−1x is, by definition, in [-pi/2,pi/2][−π2,π2] so cos(sin^-1x) >= 0cos(sin−1x)≥0

so cos(sin^-1x) = sqrt(1-x^2)cos(sin−1x)=√1−x2

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