Inverse Trigonometric Functions
Running the ratios backwards — from a value to the angle that produced it, once we agree on which angle to pick
- Why the ratios must be restricted before they can be inverted, and what a principal value is.
- The domain, range and graph of each of the six inverse functions.
- The self-composition rules — when \(\sin^{-1}(\sin\theta)=\theta\) and when it does not.
- The negative-argument, reciprocal and complementary identities.
- How to interconvert between the inverse functions.
- The addition and multiple-angle formulae — and the conditions that guard them.
Why Restrict the Domain
A function can be inverted only if it is one-to-one. But \(\sin\theta\) takes the value \(\tfrac12\) at \(\tfrac{\pi}{6}\), at \(\tfrac{5\pi}{6}\), and at infinitely many more angles — far from one-to-one. To recover a genuine inverse we agree to return just one of those angles, the principal value, by restricting each ratio to an interval on which it rises or falls without repeating.
Domains & Principal Ranges
These six rows are the foundation of the entire chapter; every later identity respects them.
| Function | Domain | Principal range |
|---|---|---|
| \(\sin^{-1}x\) | \([-1,1]\) | \(\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\) |
| \(\cos^{-1}x\) | \([-1,1]\) | \([0,\pi]\) |
| \(\tan^{-1}x\) | \(\mathbb{R}\) | \(\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)\) |
| \(\cot^{-1}x\) | \(\mathbb{R}\) | \((0,\pi)\) |
| \(\sec^{-1}x\) | \(|x|\ge1\) | \([0,\pi]\setminus\left\{\tfrac{\pi}{2}\right\}\) |
| \(\csc^{-1}x\) | \(|x|\ge1\) | \(\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\setminus\{0\}\) |
The Graphs
Each inverse graph is the reflection of the restricted ratio in the line \(y=x\). The arcsine rises across \([-1,1]\); the arccosine falls over the same domain; the arctangent climbs over all of \(\mathbb{R}\) between two horizontal asymptotes.
Self-Composition
Feeding a function into its own inverse almost cancels — but only when the inner angle already sits inside the principal range.
\(\sin\!\left(\sin^{-1}x\right)=x\) for every \(x\in[-1,1]\) — clean and unconditional. But \(\sin^{-1}(\sin\theta)=\theta\) only when \(\theta\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\). For a \(\theta\) outside that band, first reduce it to an equivalent angle inside the principal range using allied-angle rules, then read off the answer.
Negative Arguments & Reciprocals
The "sine family" is odd; the "cosine family" reflects through \(\pi\). Reciprocal arguments swap a function for its co-named partner.
| Negative argument | Reciprocal argument |
|---|---|
| \(\sin^{-1}(-x)=-\sin^{-1}x\) | \(\csc^{-1}x=\sin^{-1}\dfrac{1}{x},\ |x|\ge1\) |
| \(\tan^{-1}(-x)=-\tan^{-1}x\) | \(\sec^{-1}x=\cos^{-1}\dfrac{1}{x},\ |x|\ge1\) |
| \(\cos^{-1}(-x)=\pi-\cos^{-1}x\) | \(\cot^{-1}x=\tan^{-1}\dfrac{1}{x},\ x>0\) |
| \(\cot^{-1}(-x)=\pi-\cot^{-1}x\) | \(\cot^{-1}x=\pi+\tan^{-1}\dfrac{1}{x},\ x<0\) |
The Complementary Pairs
Each co-function pair adds to a right angle across its shared domain — three of the most-used identities in the chapter.
Interconversion
Any one inverse function can be rewritten as any other by drawing a right triangle whose acute angle is the given inverse. For \(x\ge0\) the whole family lines up.
Set \(\theta=\sin^{-1}x\), so \(\sin\theta=x\) and \(\cos\theta=\sqrt{1-x^2}\); every other ratio of \(\theta\) — and hence every other inverse name — follows from this single triangle. Mind the sign of the surd when \(x<0\).
Sum & Difference Formulae
The tangent addition law from Chapter 11 transfers to inverses — but because the result must land back in the principal range, a correction of \(\pm\pi\) is sometimes needed.
Skipping the condition is the single most common error here. For example \(\tan^{-1}2+\tan^{-1}3\) has \(xy=6>1\), so the bare formula gives \(\tan^{-1}(-1)=-\tfrac{\pi}{4}\) — wrong. The correct value adds \(\pi\): the answer is \(\tfrac{3\pi}{4}\).
Multiple-Angle Formulae
Setting \(y=x\) in the addition laws gives the double-angle results; pushing once more gives the triple-angle versions. Each carries its own validity window.
| Expression | Equals | Valid for |
|---|---|---|
| \(2\tan^{-1}x\) | \(\tan^{-1}\dfrac{2x}{1-x^{2}}\) | \(|x|<1\) |
| \(2\tan^{-1}x\) | \(\sin^{-1}\dfrac{2x}{1+x^{2}}\) | \(|x|\le1\) |
| \(2\tan^{-1}x\) | \(\cos^{-1}\dfrac{1-x^{2}}{1+x^{2}}\) | \(x\ge0\) |
| \(3\sin^{-1}x\) | \(\sin^{-1}(3x-4x^{3})\) | \(|x|\le\tfrac12\) |
| \(3\cos^{-1}x\) | \(\cos^{-1}(4x^{3}-3x)\) | \(\tfrac12\le x\le1\) |
Putting It to Work
Problem. Evaluate \(\sin^{-1}\!\left(\sin\dfrac{2\pi}{3}\right)\).
Solution. Since \(\tfrac{2\pi}{3}\notin\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\), reduce: \(\sin\tfrac{2\pi}{3}=\sin\!\left(\pi-\tfrac{2\pi}{3}\right)=\sin\tfrac{\pi}{3}\).
Problem. Evaluate \(\cos^{-1}\!\left(\cos\dfrac{7\pi}{6}\right)\).
Solution. Here \(\tfrac{7\pi}{6}\in[\pi,2\pi]\), where \(\cos^{-1}(\cos\theta)=2\pi-\theta\):
Problem. Prove \(\tan^{-1}\dfrac{1}{2}+\tan^{-1}\dfrac{1}{3}=\dfrac{\pi}{4}\).
Solution. Here \(xy=\tfrac16<1\), so the bare formula applies:
Problem. Find \(\sin\!\left(2\tan^{-1}\dfrac{1}{3}\right)\).
Solution. Using \(2\tan^{-1}x=\sin^{-1}\dfrac{2x}{1+x^2}\) with \(x=\tfrac13\):
Problem. Prove \(\sin^{-1}\dfrac{3}{5}+\sin^{-1}\dfrac{8}{17}=\sin^{-1}\dfrac{77}{85}\).
Solution. With \(x=\tfrac35,\ y=\tfrac{8}{17}\) (and \(x^2+y^2\le1\)), use \(x\sqrt{1-y^2}+y\sqrt{1-x^2}\):
Problem. Solve \(\tan^{-1}(x-1)+\tan^{-1}x+\tan^{-1}(x+1)=\tan^{-1}3x\).
Solution. Combine the outer two: \(\tan^{-1}(x-1)+\tan^{-1}(x+1)=\tan^{-1}\dfrac{2x}{2-x^2}\), then collect:
So \(x=0,\ \pm\tfrac12\).
Chapter Summary
\(\sin^{-1},\tan^{-1},\csc^{-1}\to\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\); \(\cos^{-1},\cot^{-1},\sec^{-1}\to[0,\pi]\).
\(\sin(\sin^{-1}x)=x\) always; \(\sin^{-1}(\sin\theta)=\theta\) only on the branch.
\(\sin^{-1}x+\cos^{-1}x=\tan^{-1}x+\cot^{-1}x=\tfrac{\pi}{2}\).
\(\tan^{-1}x+\tan^{-1}y=\tan^{-1}\tfrac{x+y}{1-xy}\) — check \(xy\lessgtr1\).
\(2\tan^{-1}x\) in three forms; triple-angle \(3\sin^{-1}x,\,3\cos^{-1}x\).
Every identity carries a validity window; verify it before applying.
Problems
Watch the principal ranges and the validity conditions. Difficulty rises down the list.
- Find the principal values of \(\sin^{-1}\!\left(-\tfrac12\right)\) and \(\cos^{-1}\!\left(-\tfrac{1}{\sqrt2}\right)\).
- Evaluate \(\tan^{-1}\!\left(\tan\dfrac{3\pi}{4}\right)\).
- Evaluate \(\sin^{-1}(\sin 5)\), where \(5\) is in radians.
- Prove \(\sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2}\).
- Prove \(\tan^{-1}\dfrac15+\tan^{-1}\dfrac17+\tan^{-1}\dfrac13+\tan^{-1}\dfrac18=\dfrac{\pi}{4}\).
- Show that \(2\tan^{-1}\dfrac13=\tan^{-1}\dfrac34\).
- Evaluate \(\cos\!\left(2\sin^{-1}\dfrac35\right)\).
- Solve \(\tan^{-1}(2x)+\tan^{-1}(3x)=\dfrac{\pi}{4}\).
- Show that \(2\tan^{-1}x=\cos^{-1}\dfrac{1-x^2}{1+x^2}\) for \(x\ge0\).
- Express \(\sin^{-1}\!\left(2x\sqrt{1-x^2}\right)\) in terms of \(\sin^{-1}x\) for \(-\tfrac{1}{\sqrt2}\le x\le\tfrac{1}{\sqrt2}\).
- Prove \(\tan^{-1}1+\tan^{-1}2+\tan^{-1}3=\pi\).
- Find all \(x\) for which \(\cos^{-1}x>\sin^{-1}x\).