Inverse Trigonometric Functions

[nycmath]

See attachment.

[Math Tutor]

A. Set [tex]y=2\sin x-1[/tex] and solve for [tex]x[/tex]: [tex]\sin x=\frac{y+1}{2}[/tex]. On [tex]-\frac{\pi}{2}\le x\le\frac{\pi}{2}[/tex] the sine is one-to-one, so you can apply [tex]\sin^{-1}[/tex] directly:

[tex]f^{-1}(x)=\sin^{-1}\left(\frac{x+1}{2}\right),\qquad -3\le x\le 1[/tex]

The domain of the inverse is just the range of the original — that's the quickest way to check you restricted things correctly.

[nycmath]

A. Set [tex]y=2\sin x-1[/tex] and solve for [tex]x[/tex]: [tex]\sin x=\frac{y+1}{2}[/tex]. On [tex]-\frac{\pi}{2}\le x\le\frac{\pi}{2}[/tex] the sine is one-to-one, so you can apply [tex]\sin^{-1}[/tex] directly:

[tex]f^{-1}(x)=\sin^{-1}\left(\frac{x+1}{2}\right),\qquad -3\le x\le 1[/tex]

The domain of the inverse is just the range of the original — that's the quickest way to check you restricted things correctly.

Great work as usual.