Inverse Trigonometric Equations
You know that $\sin(\pi/6) = 1/2$ — but what angle gives $\sin^{-1}(1/2)$? And what if the equation is $2\sin^{-1}(x) + \cos^{-1}(x) = \pi$? Inverse trig equations demand you track the principal range at every step. In this lesson you'll master the three inverse functions, their restricted domains and ranges, and a reliable method for solving equations that involve them.
Without a calculator: what is $\sin^{-1}(\sin(5\pi/6))$? Write your instinct below — $5\pi/6$, or something else? Explain your reasoning.
Every inverse trig equation comes down to two decisions: identify which inverse function is involved and check the principal range, then isolate and evaluate — while rejecting any solution outside that range.
The three principal ranges to memorise:
- $\sin^{-1}x$: range $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$, domain $[-1,1]$
- $\cos^{-1}x$: range $[0,\pi]$, domain $[-1,1]$
- $\tan^{-1}x$: range $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$, domain $\mathbb{R}$
Key facts
- Domains and ranges of $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$
- $\sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2}$ for $x \in [-1,1]$
- $\sin^{-1}(\sin\theta) = \theta$ only when $\theta \in [-\pi/2, \pi/2]$
Concepts
- Why inverse trig functions require restricted domains
- How to identify the correct principal-range equivalent of any angle
- The difference between solving for $x$ and solving for $\theta = f^{-1}(x)$
Skills
- Evaluate expressions like $\sin^{-1}(\cos\theta)$ and $\cos^{-1}(\sin\theta)$
- Solve equations involving one or more inverse trig functions
- Apply the identity $\sin^{-1}x + \cos^{-1}x = \pi/2$ to simplify equations
The hook asked for $\sin^{-1}(\sin(5\pi/6))$. Many students write $5\pi/6$ — but that is wrong. Here is why.
The principal range of $\sin^{-1}$ is $[-\pi/2, \pi/2]$. The angle $5\pi/6 \approx 2.618$ lies outside this interval, so the cancellation law does not apply directly.
Instead, find the equivalent angle in $[-\pi/2, \pi/2]$ with the same sine value:
So $\sin^{-1}(\sin(5\pi/6)) = \sin^{-1}(\sin(\pi/6)) = \pi/6$, since $\pi/6 \in [-\pi/2, \pi/2]$.
General rule: If $\theta \in (\pi/2, \pi)$, use $\sin^{-1}(\sin\theta) = \pi - \theta$. If $\theta \in (\pi, 3\pi/2)$, use $\sin^{-1}(\sin\theta) = \theta - 2\pi$ (or equivalently work modulo $2\pi$ back to the principal range).
The hook asked for $\sin^{-1}(\sin(5\pi/6))$. Many students write $5\pi/6$ — but that is wrong . Here is why.
Pause — copy the cancellation rule: $f^{-1}(f(x))=x$ ONLY when $x$ is in the restricted domain; otherwise use the range of $f^{-1}$ to find the correct output into your book.
Quick check: What is $\sin^{-1}\!\left(\sin\dfrac{2\pi}{3}\right)$?
We just saw the cancellation trap: $\sin^{-1}(\sin(5\pi/6))\neq5\pi/6$ because $5\pi/6$ lies outside $[-\pi/2,\pi/2]$; the correct answer is $\pi/6$ (the value in the range of $\sin^{-1}$ with the same sine). That raises a question: when solving an equation like $\cos^{-1}(2x-1)=\pi/3$, how do you apply the inverse function to both sides correctly? This card answers it → apply $\cos$ to both sides: $2x-1=\cos(\pi/3)=1/2$, so $x=3/4$.
To solve an equation like $\cos^{-1}(2x - 1) = \pi/3$, apply the cosine to both sides (the inverse of the inverse), then solve the resulting trig equation. Remember to check the domain of the original inverse function.
Method:
- Isolate the inverse trig expression on one side.
- Apply the corresponding trig function to both sides to "undo" the inverse.
- Solve the resulting equation for $x$.
- Check that $x$ is in the domain of the original inverse function.
Example: Solve $\cos^{-1}(2x-1) = \pi/3$.
Step 1: $2x - 1 = \cos(\pi/3) = 1/2$
Step 2: $2x = 3/2$, so $x = 3/4$
Step 3: Check — $2(3/4) - 1 = 1/2 \in [-1,1]$ ✓
To solve an equation like $\cos^{-1}(2x - 1) = \pi/3$, apply the cosine to both sides (the inverse of the inverse), then solve the resulting trig equation. Remember to check the domain of the original inverse function.
Pause — copy the equation-solving technique: apply the corresponding trig function to both sides to undo the inverse-trig, then solve the resulting equation into your book.
Did you get this? True or false: the solution to $\sin^{-1}(x) = -\pi/4$ is $x = -\dfrac{\sqrt{2}}{2}$.
Worked examples · 3 in a row, reveal as you go
Evaluate $\sin^{-1}\!\left(\cos\dfrac{\pi}{3}\right)$.
Solve $\tan^{-1}(3x - 1) = -\dfrac{\pi}{4}$.
Solve $\sin^{-1}(x) + \cos^{-1}(x) = \dfrac{\pi}{2}$. Hence find all $x$ satisfying $2\sin^{-1}(x) + \cos^{-1}(x) = \pi$.
Fill the gap: Using the identity, $\cos^{-1}(x) = \dfrac{\pi}{2}\ -$ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: $\cos^{-1}\!\left(-\dfrac{\sqrt{3}}{2}\right) = \dfrac{5\pi}{6}$.
Activities · practice with the ideas
Evaluate $\sin^{-1}\!\left(\sin\dfrac{7\pi}{6}\right)$. Show your reasoning clearly.
Solve $\cos^{-1}(3x + 1) = \dfrac{2\pi}{3}$.
Find the value of $\sin^{-1}\!\left(\cos\dfrac{2\pi}{3}\right)$.
Solve $3\sin^{-1}(x) - \cos^{-1}(x) = \pi$. Use the complementary identity.
If $\tan^{-1}(x) + \tan^{-1}(2) = \pi/4$, find $x$. (Hint: apply $\tan$ to both sides after isolating $\tan^{-1}(x)$, then use the compound angle formula.)
Odd one out: Three of these evaluations are correct. Which one is NOT?
Earlier you estimated $\sin^{-1}(\sin(5\pi/6))$. The correct answer is $\pi/6$ — because $\sin(5\pi/6) = \sin(\pi - 5\pi/6) = \sin(\pi/6)$, and $\pi/6$ lies in the principal range $[-\pi/2, \pi/2]$.
The key lesson: never cancel blindly. Always check whether the argument lies in the principal range. If it does not, find the equivalent angle that does.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Evaluate $\cos^{-1}\!\left(\cos\dfrac{5\pi}{4}\right)$. Show all working. (2 marks)
Q2. Solve $\sin^{-1}(2x) = \cos^{-1}(x)$, giving your answer in exact form. (3 marks)
Q3. Show that $\sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2}$ for $x \in [-1,1]$. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $\sin(7\pi/6) = -\sin(\pi/6) = -1/2$, and the angle in $[-\pi/2,\pi/2]$ with $\sin\theta = -1/2$ is $-\pi/6$. So $\sin^{-1}(\sin(7\pi/6)) = -\pi/6$.
2. $\cos^{-1}(3x+1) = 2\pi/3 \Rightarrow 3x+1 = \cos(2\pi/3) = -1/2 \Rightarrow 3x = -3/2 \Rightarrow x = -1/2$. Check: $3(-1/2)+1 = -1/2 \in [-1,1]$ ✓.
3. $\cos(2\pi/3) = -1/2$, so $\sin^{-1}(-1/2) = -\pi/6$.
4. Replace $\cos^{-1}x = \pi/2 - \sin^{-1}x$: $3\sin^{-1}x - (\pi/2 - \sin^{-1}x) = \pi \Rightarrow 4\sin^{-1}x = 3\pi/2 \Rightarrow \sin^{-1}x = 3\pi/8 \Rightarrow x = \sin(3\pi/8)$.
5. $\tan^{-1}x = \pi/4 - \tan^{-1}2$. Apply $\tan$: $x = \tan(\pi/4 - \tan^{-1}2) = \dfrac{1-2}{1+2} = -\dfrac{1}{3}$.
Q1 (2 marks): $5\pi/4 \notin [0,\pi]$, and $\cos(5\pi/4) = -\sqrt{2}/2$ [1]. The angle in $[0,\pi]$ with cosine $-\sqrt{2}/2$ is $3\pi/4$. So the answer is $3\pi/4$ [1].
Q2 (3 marks): Let $\sin^{-1}(2x) = \cos^{-1}(x)$. Using the identity, $\cos^{-1}(x) = \pi/2 - \sin^{-1}(x)$, so $\sin^{-1}(2x) = \pi/2 - \sin^{-1}(x)$ [1]. Apply $\sin$: $2x = \cos(\sin^{-1}(x)) = \sqrt{1-x^2}$ [1]. Square: $4x^2 = 1 - x^2 \Rightarrow 5x^2 = 1 \Rightarrow x = \pm 1/\sqrt{5}$. Check both — $x = 1/\sqrt{5}$ gives positive values consistent with original equation [1].
Q3 (3 marks): Let $\theta = \sin^{-1}x$, so $\sin\theta = x$ and $\theta \in [-\pi/2, \pi/2]$ [1]. Then $\pi/2 - \theta \in [0,\pi]$ [1]. Now $\cos(\pi/2 - \theta) = \sin\theta = x$, so $\cos^{-1}(x) = \pi/2 - \theta = \pi/2 - \sin^{-1}(x)$, giving $\sin^{-1}x + \cos^{-1}x = \pi/2$ [1].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering inverse trig questions. Lighter alternative to the boss.
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