Principal value branches

The six trigonometric functions are periodic, hence not one-one on their natural domains. To define an inverse we must choose a maximal interval on which the function is monotonic and onto its full image. The chosen interval is called the principal value branch, and the corresponding inverse is called the principal value.

The need for restriction

The sine function $\sin : \mathbb{R} \to [-1, 1]$ is onto its codomain but is far from one-one: $\sin 0 = \sin \pi = \sin 2\pi = 0$. If we tried to write $\sin^{-1} 0$, we would have infinitely many candidates. The remedy is to insist that $\sin^{-1} y$ should give the unique angle in a chosen interval. The interval is conventional, but the conventions are fixed worldwide.

The six principal branches

Each restriction is the maximal monotonic interval near zero that gives the full image. Memorise the table.

The inverse on each principal branch is denoted $\sin^{-1}, \cos^{-1}, \tan^{-1}, \cot^{-1}, \sec^{-1}, \csc^{-1}$ and called the principal value.

Why these intervals

For $\sin$ the interval $[-\pi/2, \pi/2]$ is the unique maximal interval containing $0$ on which $\sin$ is strictly increasing. Any other choice (such as $[\pi/2, 3\pi/2]$) would also work, but conventions favour intervals containing zero.

For $\cos$ the function is decreasing on $[0, \pi]$. Note that $0$ is included (not excluded) , many students worry whether the principal branch should be open or closed. It is closed where the function takes finite values.

For $\tan$ the open interval $(-\pi/2, \pi/2)$ excludes the asymptotes. For $\cot$, the open interval $(0, \pi)$ does the same.

For $\sec$ and $\csc$, the principal branches are restrictions of the cosine and sine branches with the point where $\cos$ or $\sin$ equals zero removed (because $\sec$ or $\csc$ would be undefined there).

Reading principal values

Principal value is the angle in the principal branch whose sine (cosine, etc.) equals the given value.

Examples:

• $\sin^{-1} \tfrac{1}{2}$: angle in $[-\pi/2, \pi/2]$ with sine $\tfrac{1}{2}$. Answer: $\pi/6$.
• $\sin^{-1}(-\tfrac{1}{2})$: angle in $[-\pi/2, \pi/2]$ with sine $-\tfrac{1}{2}$. Answer: $-\pi/6$.
• $\cos^{-1}(-\tfrac{\sqrt 3}{2})$: angle in $[0, \pi]$ with cosine $-\tfrac{\sqrt 3}{2}$. Answer: $5\pi/6$.
• $\tan^{-1} 1$: angle in $(-\pi/2, \pi/2)$ with tangent $1$. Answer: $\pi/4$.
• $\tan^{-1}(-1)$: angle in $(-\pi/2, \pi/2)$ with tangent $-1$. Answer: $-\pi/4$.

Identity inside the branch

If $x$ lies in the principal branch of the trig function, then $\sin^{-1}(\sin x) = x$, $\cos^{-1}(\cos x) = x$, etc. Outside the principal branch one must reduce.

Example. $\sin^{-1}(\sin \tfrac{5\pi}{6})$. The angle $\tfrac{5\pi}{6}$ is not in $[-\pi/2, \pi/2]$. But $\sin \tfrac{5\pi}{6} = \sin(\pi - \tfrac{5\pi}{6}) = \sin \tfrac{\pi}{6}$, and $\tfrac{\pi}{6}$ is in the principal branch. So the answer is $\tfrac{\pi}{6}$.

Example. $\cos^{-1}(\cos \tfrac{7\pi}{6})$. The angle $\tfrac{7\pi}{6}$ is not in $[0, \pi]$. But $\cos \tfrac{7\pi}{6} = \cos(2\pi - \tfrac{7\pi}{6}) = \cos \tfrac{5\pi}{6}$ and $\tfrac{5\pi}{6} \in [0, \pi]$. So the answer is $\tfrac{5\pi}{6}$.

Worked examples

Example 1. Evaluate $\sin^{-1}(-\tfrac{\sqrt 3}{2})$.

Looking for $\theta \in [-\pi/2, \pi/2]$ with $\sin \theta = -\tfrac{\sqrt 3}{2}$. $\theta = -\pi/3$.

Example 2. Evaluate $\cos^{-1}(\cos(-\tfrac{\pi}{4}))$.

$\cos$ is even, so $\cos(-\pi/4) = \cos(\pi/4) = \tfrac{\sqrt 2}{2}$. Then $\cos^{-1}(\tfrac{\sqrt 2}{2}) = \pi/4$. Note the answer is not $-\pi/4$ because $\cos^{-1}$ outputs in $[0, \pi]$.

Example 3. Evaluate $\tan^{-1}(\tan \tfrac{3\pi}{4})$.

$\tfrac{3\pi}{4}$ is not in $(-\pi/2, \pi/2)$. Subtract $\pi$: $\tan(\tfrac{3\pi}{4}) = \tan(-\tfrac{\pi}{4}) = -1$. So $\tan^{-1}(-1) = -\pi/4$.

Example 4. Evaluate $\sin^{-1}(\sin 10)$ in radians.

Reduce $10$ modulo $2\pi$: $10 - 2\pi \approx 10 - 6.283 \approx 3.717$, still outside $[-\pi/2, \pi/2]$. Note $\sin(\pi - 3.717) = \sin 3.717$, and $\pi - 3.717 \approx -0.576$, in the principal branch. So $\sin^{-1}(\sin 10) = \pi - 10 + 2\pi = 3\pi - 10 \approx -0.575$. (Alternatively: $10 - 3\pi \approx 0.575$, but the sign is negative because of how we matched signs.)

Careful: $\sin^{-1}(\sin 10)$ equals $3\pi - 10$ if that lies in $[-\pi/2, \pi/2]$. Check: $3\pi - 10 \approx -0.575$, which is in $[-\pi/2, \pi/2] \approx [-1.571, 1.571]$. Yes. Answer $3\pi - 10$.

Example 5. State the domain and range of $\sec^{-1} x$.

Domain: $\mathbb{R} \setminus (-1, 1)$, i.e., $|x| \ge 1$. Range: $[0, \pi] \setminus \{\pi/2\}$.

Example 6. $\sin^{-1}(\sin(-\tfrac{17\pi}{8}))$.

Add $2\pi$: $-\tfrac{17\pi}{8} + 2\pi = -\tfrac{\pi}{8} \in [-\pi/2, \pi/2]$. So the answer is $-\pi/8$.

Try it yourself

1. $\sin^{-1}(\tfrac{\sqrt 3}{2}) = ?$
2. $\cos^{-1}(0) = ?$
3. $\tan^{-1}(\sqrt 3) = ?$
4. $\cot^{-1}(0) = ?$
5. $\sec^{-1}(-2) = ?$
6. $\sin^{-1}(\sin \tfrac{3\pi}{4}) = ?$
7. $\cos^{-1}(\cos(-\tfrac{\pi}{6})) = ?$
8. $\tan^{-1}(\tan \tfrac{5\pi}{6}) = ?$
9. State why $\sin^{-1}(\sin x) \neq x$ for $x = \pi$.
10. Find the value of $\csc^{-1}(\sqrt 2)$.
11. Find $\sin^{-1}(\sin 8)$ in radians (use $\pi \approx 3.14$).
12. Sketch the graph of $\sin^{-1} x$.
13. State the range of $\arccos$.
14. If $\sin^{-1} x = \pi/3$, find $x$ and then $\cos^{-1} x$.

Pitfalls / Tricks

• Always check the principal range first; an answer outside that range is wrong.
• $\sin^{-1}(\sin x) = x$ only when $x \in [-\pi/2, \pi/2]$. Outside this range, reduce.
• $\sin^{-1}$ is not the reciprocal $1/\sin$; that one would be $\csc$.
• The inverse cosine of a negative number lies in $(\pi/2, \pi]$, not in $(-\pi/2, 0)$.
• Conventions for $\sec^{-1}$ and $\csc^{-1}$ vary between textbooks; we follow the standard Indian board convention.

The next subtopic plots these inverses and discusses symmetry.