Equations & Inequalities with Inverse & Reciprocal Trigonometric Functions (College Board AP® Precalculus): Study Guide

Equations & inequalities with arcsin, arccos and arctan

How can I solve equations involving inverse trigonometric functions?

• An equation of the form ${\sin }^{-1}\left(\text{expression}\right)=c$ (or similar with ${\cos }^{-1}$ or ${\tan }^{-1}$) can be solved by applying the corresponding trigonometric function to both sides

  • For example, given ${\sin }^{-1}\left(x\right)=c$, applying sine to both sides gives:

    • $\sin \left({\sin }^{-1}\right(x\left)\right)=\sin c\Rightarrow x=\sin c$

  • This converts the inverse trig equation into a simple trig equation

• In general:

  • Start by applying the appropriate trigonometric function to both sides to remove the inverse trig function

  • Then solve the resulting equation

  • Check that the solution is in the domain of the original inverse trig function

    • i.e. that the solution is a valid input for the inverse trig function

      • in the interval $\left[-1,1\right]$ for ${\sin }^{-1}$ and ${\cos }^{-1}$

      • or any real number for ${\tan }^{-1}$

  • Also check that the value of $c$ on the right-hand side of the original equation is in the range of the inverse trig function

    • i.e. $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ for ${\sin }^{-1}$, $[0,\pi ]$ for ${\cos }^{-1}$, $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ for ${\tan }^{-1}$

    • If $c$ is outside this range, the equation has no solution

How can I solve inequalities involving inverse trigonometric functions?

• Note these important behaviors of the three inverse trigonometric functions:

  • ${\sin }^{-1}$ and ${\tan }^{-1}$ are always increasing everywhere on their domains

  • ${\cos }^{-1}$ is always decreasing everywhere on its domain

• This makes inverse trig inequalities relatively straightforward to solve

  • Start by solving the corresponding equation (replacing the inequality sign with $=$)

    • This will give you the boundary value

  • Then use the behavior of the inverse trig function to determine which side of the boundary satisfies the inequality

  • Check the domain of the inverse trig function

    • for ${\sin }^{-1}$ and ${\cos }^{-1}$, the expression inside must lie in $[-1,1]$

    • for ${\tan }^{-1}$, any real number is allowed

  • Combine the inequality with the domain restriction to give the final solution

• E.g. to solve ${\sin }^{-1}(2x-1)>\frac{\pi }{6}$

  • The corresponding equation is ${\sin }^{-1}(2x-1)=\frac{\pi }{6}$

    • $\begin{array}{rcl}2x-1& =& \sin \frac{\pi }{6}\Rightarrow 2x-1=\frac{1}{2}\Rightarrow x=\frac{3}{4}\end{array}$

  • Since ${\sin }^{-1}$ is increasing, ${\sin }^{-1}(2x-1)$ will be greater than $\frac{\pi }{6}$ to the right of $x=\frac{3}{4}$

    • This gives $x>\frac{3}{4}$

  • But the domain restriction is

    • $-1\le 2x-1\le 1\Rightarrow 0\le x\le 1$

  • Combining these gives the final solution

    • $\frac{3}{4}<x\le 1$

Equations & inequalities with sec, csc and cot

How can I solve reciprocal trigonometric equations?

• A simple reciprocal trigonometric equation is one of the form

  • $\sec \theta =k,\csc \theta =k,\text{or}\cot \theta =k$

    • where $k$ is a constant

• More generally, a question may give an equation that simplifies to one of these forms after some algebraic rearrangement

  • e.g. $3\sec \theta +1=7$, which can be rearranged to give $\sec \theta =2$

• The goal is to find all values of $\theta$ in a specified solution interval that make the equation true

• The most reliable approach is to convert the equation into an equation involving sine, cosine, or tangent

  • by using the reciprocal relationships

    • $\sec \theta =\frac{1}{\cos \theta },\csc \theta =\frac{1}{\sin \theta },\cot \theta =\frac{1}{\tan \theta }$

  • Once converted, the equation can be solved using the methods for simple trigonometric equations

• In general:

  • Start by rearranging the equation to isolate the reciprocal trig function on one side

    • i.e. in the form $\sec \theta =k$, $\csc \theta =k$, or $\cot \theta =k$

  • Then take the reciprocal of both sides to convert to an equation involving cosine, sine, or tangent

    • $\sec \theta =k$ becomes $\cos \theta =\frac{1}{k}$

    • $\csc \theta =k$ becomes $\sin \theta =\frac{1}{k}$

    • $\cot \theta =k$ becomes $\tan \theta =\frac{1}{k}$

  • Solve the resulting equation using the methods for simple trigonometric equations

    • i.e. find an initial solution, then use symmetry and periodicity to find all other solutions in the interval

When does a reciprocal trigonometric equation have no solution?

• The secant and cosecant functions have a range of $(-\infty , -1]\cup [1, \infty )$

  • i.e. they never take values in the interval $\left(-1,1\right)$

• Therefore, equations like $\sec \theta =0.5$ or $\csc \theta =-0.3$ have no solution

  • because the right-hand side is not in the range of the function

• This can also be seen from the reciprocal step

  • E.g. $\sec \theta =0.5$ would give $\cos \theta =\frac{1}{0.5}=2$

    • but cosine cannot exceed $1$

• The cotangent function outputs all real number values

  • so $\cot \theta =k$ always has solutions

How can I solve inequalities with reciprocal trigonometric functions?

• A reciprocal trigonometric inequality has the form $\sec \theta <k$, $\csc \theta \ge k$, $\cot \theta >k$, etc.

• The general approach is similar to solving the equations

  • Start by rearranging to isolate the reciprocal trig function

  • Solve the corresponding equation (with $=$ instead of the inequality sign)

    • This will give you the boundary values

  • Identify which intervals satisfy the inequality

    • E.g. by using a graph or by testing values from each interval

• Take particular care with reciprocal trig inequalities because the reciprocal functions have vertical asymptotes

  • These split the solution intervals

  • E.g. when solving an inequality involving secant

    • the asymptotes at $\theta =\frac{\pi }{2}+k\pi$ may divide the solution set into multiple disjoint intervals