For 0&lt;x&lt;1 express sin⁡[sin−1(x)+cos−1(x)], in terms of x sin⁡[sin−1(x)+cos−1(x)] sin⁡[sin−1(x)]⋅cos⁡[cos−1(x)]+cos⁡[sin−1(x)]⋅sin⁡[cos−1(x)] x⋅x+cos⁡[sin−1(x)]⋅sin⁡[cos−1(x)] x2+cos⁡[sin−1(x)]⋅sin⁡[cos−1(x)] Thats

Using the fact cos⁡(u)=1−sin2u sin⁡(u)=1−cos2u we have cos⁡[sin−1x]⋅sin⁡[cos−1x]=1−sin2(sin−1x)⋅1−cos2(cos−1x)=1−x2 therefore sin⁡[sin−1x+cos−1x]=1

"For 0<x<1 express sin⁡[sin−1(x)+cos−1(x)], in terms of x..." was answered by our community

Cameron Russell

2022-01-28

For

0 < x < 1

express

\sin [\sin^{- 1} (x) + \cos^{- 1} (x)]

, in terms of x

\sin [\sin^{- 1} (x) + \cos^{- 1} (x)]

\sin [\sin^{- 1} (x)] \cdot \cos [\cos^{- 1} (x)] + \cos [\sin^{- 1} (x)] \cdot \sin [\cos^{- 1} (x)]

x \cdot x + \cos [\sin^{- 1} (x)] \cdot \sin [\cos^{- 1} (x)]

x^{2} + \cos [\sin^{- 1} (x)] \cdot \sin [\cos^{- 1} (x)]

Thats

spelkw

Beginner2022-01-29Added 12 answers

Using the fact

\cos (u) = \sqrt{1 - \sin^{2} u}

\sin (u) = \sqrt{1 - \cos^{2} u}

we have

\cos [\sin^{- 1} x] \cdot \sin [\cos^{- 1} x] = \sqrt{1 - \sin^{2} (\sin^{- 1} x)} \cdot \sqrt{1 - \cos^{2} (\cos^{- 1} x)} = 1 - x^{2}

therefore

\sin [\sin^{- 1} x + \cos^{- 1} x] = 1

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