Using the table of values for $f\left(x\right)=\sin x$f(x)=sinx on the domain $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$[−π2,π2], complete the table of values for $f^{-1}\left(x\right)=\sin^{-1}\left(x\right)$f−1(x)=sin−1(x).
$x$x | $-\frac{\pi}{2}$−π2 | $-\frac{\pi}{4}$−π4 | $0$0 | $\frac{\pi}{4}$π4 | $\frac{\pi}{2}$π2 |
---|---|---|---|---|---|
$f\left(x\right)=\sin x$f(x)=sinx | $-1$−1 | $-\frac{\sqrt{2}}{2}$−√22 | $0$0 | $\frac{\sqrt{2}}{2}$√22 | $1$1 |
$x$x | $-1$−1 | $-\frac{\sqrt{2}}{2}$−√22 | $0$0 | $\frac{\sqrt{2}}{2}$√22 | $1$1 |
---|---|---|---|---|---|
$f^{-1}\left(x\right)=\sin^{-1}\left(x\right)$f−1(x)=sin−1(x) | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ |
Using the table of values for $f\left(x\right)=\cos x$f(x)=cosx on the domain $\left[0,\pi\right]$[0,π], complete the table of values for $f^{-1}\left(x\right)=\cos^{-1}\left(x\right)$f−1(x)=cos−1(x).
Using the table of values for $f\left(x\right)=\tan x$f(x)=tanx on the domain $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$(−π2,π2), complete the table of values for $f^{-1}\left(x\right)=\tan^{-1}\left(x\right)$f−1(x)=tan−1(x).
Consider that $\sin\left(\frac{\pi}{6}\right)=\frac{1}{2}$sin(π6)=12.