Question

2 sin−1(x) + cos−1(x) = π
The hint is to take the cosine of each side

Answer 1

First take
`y=\sin^(-1)x` and
`z=\cos^(-1)x`. Then


`\cos(2\sin^(-1)x+\cos^(-1)x)=\cos(2y+z)=\cos2y\cos z-\sin2y\sin z`

`=(\cos^2y-\sin^2y)\cos z-2\sin y\cos y\sin z`

Now


`\sin y=\sin(\sin^(-1)x)=x`

`\cos y=\cos(\sin^(-1)x)=√(1-x^2)`

both provided that
`-\frac\pi2\le y\le\frac\pi2`, and


`\sin z=\sin(\cos^(-1)x)=√(1-x^2)`

`\cos z=\cos(\cos^(-1)x)=x`

both provided that
`0\le z\le\pi`.

Then


`\cos(2\sin^(-1)x+\cos^(-1)x)=(1-2x^2)x-2x(1-x^2)`

`\implies -x=\cos\pi\implies -x=-1\implies x=1`