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7.02 Inverse trigonometric functions

Interactive practice questions

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 $f^{-1}\left(x\right)=\sin^{-1}\left(x\right)$f1(x)=sin1(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
$f^{-1}\left(x\right)=\sin^{-1}\left(x\right)$f1(x)=sin1(x) $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
Easy
2min

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 $f^{-1}\left(x\right)=\cos^{-1}\left(x\right)$f1(x)=cos1(x).

Easy
1min

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 $f^{-1}\left(x\right)=\tan^{-1}\left(x\right)$f1(x)=tan1(x).

Easy
< 1min

Consider that $\sin\left(\frac{\pi}{6}\right)=\frac{1}{2}$sin(π6)=12.

Easy
< 1min
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