Inverse Trigonometric Functions (DP IB Analysis & Approaches (AA): HL): Revision Note

Inverse trig functions

What are the inverse trig functions?

• The inverse trig functions are

  • $\arcsin x={\sin }^{-1}x$

  • $\mathrm{arc}\cos x={\cos }^{-1}x$

  • $\mathrm{arc}\tan x={\tan }^{-1}x$

• They are inverses when the domains are restricted

  • $\arcsin x$ is the inverse of $\sin x$ when its domain is $-\frac{\mathrm{\pi }}{2}\le x\le \frac{\mathrm{\pi }}{2}$

    • e.g. $\sin \left(\frac{\mathrm{\pi }}{2}\right)=1\Rightarrow \arcsin \left(1\right)=\frac{\mathrm{\pi }}{2}$

  • $\mathrm{arc}\cos x$ is the inverse of $\cos x$ when its domain is $0\le x\le \mathrm{\pi }$

    • e.g. $\cos \left(\frac{\mathrm{\pi }}{3}\right)=\frac{1}{2}\Rightarrow \mathrm{arc}\cos \left(\frac{1}{2}\right)=\frac{\mathrm{\pi }}{3}$

  • $\mathrm{arc}\tan x$ is the inverse of $\tan x$ when its domain is $-\frac{\mathrm{\pi }}{2}<x<\frac{\mathrm{\pi }}{2}$

    • e.g. $\tan \left(\frac{\mathrm{\pi }}{4}\right)=1\Rightarrow \mathrm{arc}\tan \left(1\right)=\frac{\mathrm{\pi }}{4}$

What do the graphs of the inverse trig functions look like?

y=arcsinx

• $y=\arcsin x$ is a reflection of $y=\sin x$ in the line $y=x$ when its domain is $-\frac{\mathrm{\pi }}{2}\le x\le \frac{\mathrm{\pi }}{2}$

• The domain is $-1\le x\le 1$

• The range is $-\frac{\pi }{2}\le y\le \frac{\pi }{2}$

y=arccosx

• $y=\mathrm{arc}\cos x$ is a reflection of $y=\cos x$ in the line $y=x$ when its domain is $0\le x\le \mathrm{\pi }$

• The domain is $-1\le x\le 1$

• The range is $0\le y\le \pi$

y=arctanx

• $y=\mathrm{arc}\tan x$ is a reflection of $y=\tan x$ in the line $y=x$ when its domain is $-\frac{\mathrm{\pi }}{2}<x<\frac{\mathrm{\pi }}{2}$

• The domain is $x\in \mathrm{ℝ}$

• The range is $-\frac{\pi }{2}<y<\frac{\pi }{2}$

• There are horizontal asymptotes at $y=\pm \frac{\mathrm{\pi }}{2}$

How can I use inverse trig functions?

• You can use the inverse trig functions to help you to solve trig equations

  • e.g. if $\sin x=\frac{1}{3}$ then one solution is $x=\arcsin \left(\frac{1}{3}\right)$

  • You need to use the symmetries of the trig functions to find the other solutions

• You can solve equations involving inverse trig equations

  • $\arcsin x=k\Rightarrow x=\sin k$

  • The converse is not always true

  • You need to use the symmetries to make the value in the trig function be within its restricted domain

    • e.g. $\sin \left(\frac{7\mathrm{\pi }}{8}\right)=\sin \left(\frac{\mathrm{\pi }}{8}\right)$

    • Therefore $\sin \left(\frac{7\mathrm{\pi }}{8}\right)=x\Rightarrow \arcsin x=\frac{\mathrm{\pi }}{8}$