Basic Concepts and Principal Value Branches

The problem: trigonometric functions are not invertible

From Chapter 1, only a one-one and onto function has an inverse. The six trigonometric functions fail spectacularly: $\sin : \mathbb{R} \to [-1, 1]$ repeats its values endlessly ($\sin 0 = \sin \pi = \sin 2\pi = 0$), so it is many-one — no inverse exists on the full domain.

The fix is the same trick that rescued $x^2$: restrict the domain until the function becomes bijective. Restricted to $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, the sine function climbs steadily from $-1$ to $1$, hitting each value exactly once — one-one and onto $[-1, 1]$, hence invertible.

Branches and the principal value branch

Sine restricted to $\left[-\frac{3\pi}{2}, -\frac{\pi}{2}\right]$, or $\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$, or any such interval is also bijective onto $[-1, 1]$ — each choice produces a different branch of the inverse. To make $\sin^{-1}$ a single well-defined function, one branch is fixed as the principal value branch:

$$\sin^{-1} : [-1, 1] \to \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$

The same story, told six times, gives the whole table:

How to remember the ranges: $\sin^{-1}, \tan^{-1}, \mathrm{cosec}^{-1}$ (the "odd" trio) live in the symmetric window around 0; $\ \cos^{-1}, \cot^{-1}, \sec^{-1}$ live in the window $[0, \pi]$. Open vs closed endpoints follow the original function: $\tan$ blows up at $\pm\frac{\pi}{2}$ (open), $\sec$ is undefined at $\frac{\pi}{2}$ (deleted point).

Graphs: reflection in $y = x$

As for any inverse, the graph of $y = \sin^{-1} x$ is the mirror image of the restricted $y = \sin x$ in the line $y = x$ — interchange the axes and the point $(a, b)$ becomes $(b, a)$. The figure shows the principal branch (dark) and its reflection.

Notation warning (a guaranteed 1-mark trap): $\sin^{-1} x$ is the inverse function, NOT $\frac{1}{\sin x}$. The reciprocal is written $(\sin x)^{-1}$. The alternative name $\arcsin x$ avoids the ambiguity entirely.

Computing Principal Values

The template

The principal value of an inverse trigonometric expression is the one value lying in the principal range. To find $\sin^{-1}\left(-\frac{1}{2}\right)$:

Step 1 — set the equation: let $y = \sin^{-1}\left(-\frac{1}{2}\right)$, so $\sin y = -\frac{1}{2}$.

Step 2 — recall the reference angle: $\sin\frac{\pi}{6} = \frac{1}{2}$, so $\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}$.

Step 3 — check the branch: $-\frac{\pi}{6} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ ✓.

Answer: $\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$. Among the infinitely many angles whose sine is $-\frac{1}{2}$, the principal branch selects exactly this one.

Negative inputs: where each function puts them

The branch geometry decides how minus signs come out:

$$\sin^{-1}, \tan^{-1}, \mathrm{cosec}^{-1} \text{ of a negative number are } \textbf{negative} \ \left(\text{in } \left[-\tfrac{\pi}{2}, 0\right)\right)$$ $$\cos^{-1}, \cot^{-1}, \sec^{-1} \text{ of a negative number lie in } \left(\tfrac{\pi}{2}, \pi\right] \ \textbf{(second quadrant)}$$

So $\tan^{-1}(-1) = -\frac{\pi}{4}$ but $\cos^{-1}\left(-\frac{1}{2}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ — the odd trio flips the sign, the even trio subtracts from $\pi$.

Worked principal values (the standard set)

$$\cos^{-1}\frac{\sqrt 3}{2} = \frac{\pi}{6}, \qquad \mathrm{cosec}^{-1}(2) = \frac{\pi}{6}, \qquad \tan^{-1}(-\sqrt 3) = -\frac{\pi}{3}$$ $$\sec^{-1}\frac{2}{\sqrt 3} = \frac{\pi}{6}, \qquad \cot^{-1}(\sqrt 3) = \frac{\pi}{6}, \qquad \cos^{-1}\left(-\frac{1}{\sqrt 2}\right) = \frac{3\pi}{4}, \qquad \mathrm{cosec}^{-1}(-\sqrt 2) = -\frac{\pi}{4}$$

For $\mathrm{cosec}^{-1}$ and $\sec^{-1}$, convert to the reciprocal first if it helps: $\mathrm{cosec}^{-1}(2)$ asks for the angle whose cosecant is 2, i.e. whose sine is $\frac{1}{2}$.

Combining principal values

Sums of principal values are evaluated term by term: $$\tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right) = \frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6} = \frac{3\pi + 8\pi - 2\pi}{12} = \frac{3\pi}{4}$$ $$\cos^{-1}\frac{1}{2} + 2\sin^{-1}\frac{1}{2} = \frac{\pi}{3} + 2 \cdot \frac{\pi}{6} = \frac{2\pi}{3}$$

Common mistakes to avoid

Mistake 1 — giving a non-principal answer. $\sin^{-1}\frac{1}{2} = \frac{5\pi}{6}$? No: $\frac{5\pi}{6}$ has the right sine but lies outside $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Only $\frac{\pi}{6}$ qualifies.

Mistake 2 — flipping the sign for $\cos^{-1}$. $\cos^{-1}\left(-\frac{1}{2}\right)$ is NOT $-\frac{\pi}{3}$ — the range $[0, \pi]$ contains no negative angles. It is $\pi - \frac{\pi}{3} = \frac{2\pi}{3}$.

Mistake 3 — reading $\sin^{-1} x$ as $\frac{1}{\sin x}$. The notation trap; see the warning above.

Mistake 4 — domain slips. $\sin^{-1}(2)$ does not exist ($2 \notin [-1, 1]$), and $\sec^{-1}\left(\frac{1}{2}\right)$ does not exist either ($\frac{1}{2}$ lies in the excluded gap $(-1, 1)$).

Solved Examples

Example 1 — The model principal value

Find the principal value of $\sin^{-1}\left(\frac{1}{\sqrt 2}\right)$.

Step 1 — set up: let $y = \sin^{-1}\frac{1}{\sqrt 2}$, so $\sin y = \frac{1}{\sqrt 2}$.

Step 2 — reference angle: $\sin\frac{\pi}{4} = \frac{1}{\sqrt 2}$, and $\frac{\pi}{4} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ ✓.

Answer: $\dfrac{\pi}{4}$.

Example 2 — A second-quadrant cotangent

Find the principal value of $\cot^{-1}\left(-\frac{1}{\sqrt 3}\right)$.

Step 1 — set up: $\cot y = -\frac{1}{\sqrt 3}$ with $y$ required in $(0, \pi)$.

Step 2 — negative cotangent puts $y$ in the second quadrant: $\cot\frac{\pi}{3} = \frac{1}{\sqrt 3}$, so $$y = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$$

Step 3 — check: $\frac{2\pi}{3} \in (0, \pi)$ ✓.

Answer: $\dfrac{2\pi}{3}$ — for the $[0, \pi]$-family, negative inputs mean "subtract the reference angle from $\pi$", never "attach a minus sign".

Example 3 — The odd trio with negatives

Find the principal values of (i) $\tan^{-1}(-\sqrt 3)$, (ii) $\mathrm{cosec}^{-1}(-\sqrt 2)$.

Step 1 — (i): $\tan\frac{\pi}{3} = \sqrt 3$, and $\tan^{-1}$ of a negative is negative: $-\frac{\pi}{3} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ ✓.

Step 2 — (ii): $\mathrm{cosec}\,\frac{\pi}{4} = \sqrt 2$, so the answer is $-\frac{\pi}{4}$, legal in the cosecant branch.

Answer: (i) $-\dfrac{\pi}{3}$, (ii) $-\dfrac{\pi}{4}$.

Example 4 — Reciprocal conversion for sec and cosec

Find the principal value of $\sec^{-1}\left(\frac{2}{\sqrt 3}\right)$.

Step 1 — convert: $\sec y = \frac{2}{\sqrt 3}$ means $\cos y = \frac{\sqrt 3}{2}$.

Step 2 — reference angle: $\cos\frac{\pi}{6} = \frac{\sqrt 3}{2}$, and $\frac{\pi}{6} \in [0, \pi] - \left\{\frac{\pi}{2}\right\}$ ✓.

Answer: $\dfrac{\pi}{6}$.

Example 5 — A three-term combination

Evaluate $\tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right)$.

Step 1 — each principal value separately: $$\tan^{-1}(1) = \frac{\pi}{4}, \qquad \cos^{-1}\left(-\frac{1}{2}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}, \qquad \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$$

Step 2 — add over denominator 12: $$\frac{3\pi}{12} + \frac{8\pi}{12} - \frac{2\pi}{12} = \frac{9\pi}{12} = \frac{3\pi}{4}$$

Answer: $\dfrac{3\pi}{4}$ — note how the same input $-\frac{1}{2}$ produced a second-quadrant answer under $\cos^{-1}$ and a negative answer under $\sin^{-1}$.

Example 6 — Another combination

Evaluate $\cos^{-1}\left(\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right)$.

Step 1 — principal values: $\cos^{-1}\frac{1}{2} = \frac{\pi}{3}$ and $\sin^{-1}\frac{1}{2} = \frac{\pi}{6}$.

Step 2 — combine: $\frac{\pi}{3} + 2 \cdot \frac{\pi}{6} = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3}$.

Answer: $\dfrac{2\pi}{3}$.

Example 7 — Reading domains

Which of the following exist? (i) $\sin^{-1}(1.5)$, (ii) $\sec^{-1}(0.8)$, (iii) $\tan^{-1}(1000)$, (iv) $\cot^{-1}(-5)$.

Step 1 — check each against the domain table: (i) $1.5 \notin [-1, 1]$ — does not exist. (ii) $0.8 \in (-1, 1)$, the excluded gap for $\sec^{-1}$ — does not exist. (iii) $\tan^{-1}$ accepts all reals — exists (a value just under $\frac{\pi}{2}$). (iv) $\cot^{-1}$ accepts all reals — exists (a second-quadrant value).

Answer: only (iii) and (iv) exist. The domains $[-1,1]$ (for $\sin^{-1}, \cos^{-1}$) and $\mathbb{R} - (-1, 1)$ (for $\sec^{-1}, \mathrm{cosec}^{-1}$) are complementary — a number in the open interval $(-1,1)$ feeds the first pair only, a number beyond it feeds the second pair only, and $\pm 1$ feed all four.