NZ Level 8 (NZC) Level 3 (NCEA) [In development]
Intro to Inverse Trig Functions

Interactive practice questions

Consider the equation $\cos x=0.4$cosx=0.4, where $0\le x\le\pi$0xπ.

Which of the following could be a correct step to solve for $x$x?

Select all that apply.

$\arccos\left(\cos x\right)=\arccos\left(0.4\right)$arccos(cosx)=arccos(0.4)

A

$\frac{\cos x}{\cos}=\frac{0.4}{\cos}$cosxcos=0.4cos

B

$\cos\left(\frac{x}{x}\right)=\frac{0.4}{x}$cos(xx)=0.4x

C

$\cos^{-1}\left(\cos x\right)=\cos^{-1}\left(0.4\right)$cos1(cosx)=cos1(0.4)

D

$\arccos\left(\cos x\right)=\arccos\left(0.4\right)$arccos(cosx)=arccos(0.4)

A

$\frac{\cos x}{\cos}=\frac{0.4}{\cos}$cosxcos=0.4cos

B

$\cos\left(\frac{x}{x}\right)=\frac{0.4}{x}$cos(xx)=0.4x

C

$\cos^{-1}\left(\cos x\right)=\cos^{-1}\left(0.4\right)$cos1(cosx)=cos1(0.4)

D
Easy
Approx a minute

Consider the equation $\sin x=0.4$sinx=0.4, where $-\frac{\pi}{2}\le x\le\frac{\pi}{2}$π2xπ2.

Which of the following could be a correct step to solve for $x$x?

Select all that apply.

Consider the equation $\tan x=0.3$tanx=0.3, where $-\frac{\pi}{2}\le x\le\frac{\pi}{2}$π2xπ2.

Which of the following could be a correct step to solve for $x$x?

Select all that apply.

Consider the equations $\cos x=\frac{1}{\sqrt{2}}$cosx=12 and $x=\cos^{-1}\left(\frac{1}{\sqrt{2}}\right)$x=cos1(12).

Outcomes

M8-2

Display and interpret the graphs of functions with the graphs of their inverse and/or reciprocal functions