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:
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:
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)
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].
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