Inverse trigonometric function graphs — domain, range, principal value selection

Question

Find the principal value of $\sin^{-1}(-1/2)$ and $\cos^{-1}(-\sqrt{3}/2)$ . Why are the domains and ranges restricted for inverse trig functions? State the domain and range for all six inverse trig functions.

(CBSE 12 & JEE Main — tested every year)

Solution — Step by Step

$\sin\theta = 1/2$ has infinitely many solutions: $30°, 150°, 390°, \ldots$ But a function must give exactly ONE output for each input. So we restrict the range to give a unique answer — this is the principal value branch.

Range of $\sin^{-1}$ is $[-\pi/2, \pi/2]$ .

We need $\theta \in [-\pi/2, \pi/2]$ such that $\sin\theta = -1/2$ .

Since sine is negative and we’re in $[-\pi/2, \pi/2]$ , the angle is in the fourth quadrant (negative part): $\theta = -\pi/6$ .

\sin^{-1}(-1/2) = \mathbf{-\frac{\pi}{6}}

Range of $\cos^{-1}$ is $[0, \pi]$ .

We need $\theta \in [0, \pi]$ such that $\cos\theta = -\sqrt{3}/2$ .

Cosine is negative in the second quadrant: $\theta = \pi - \pi/6 = 5\pi/6$ .

\cos^{-1}(-\sqrt{3}/2) = \mathbf{\frac{5\pi}{6}}

Function	Domain	Range (Principal Value)
$\sin^{-1}x$	$[-1, 1]$	$[-\pi/2, \pi/2]$
$\cos^{-1}x$	$[-1, 1]$	$[0, \pi]$
$\tan^{-1}x$	$(-\infty, \infty)$	$(-\pi/2, \pi/2)$
$\cot^{-1}x$	$(-\infty, \infty)$	$(0, \pi)$
$\sec^{-1}x$	$(-\infty, -1] \cup [1, \infty)$	$[0, \pi] - \{\pi/2\}$
$\csc^{-1}x$	$(-\infty, -1] \cup [1, \infty)$	$[-\pi/2, \pi/2] - \{0\}$

Why This Works

The restriction makes each trig function one-to-one on the chosen interval, guaranteeing a unique inverse. The principal value branches are chosen by convention — they cover all possible output values exactly once.

graph TD
    A["Evaluate inverse trig"] --> B["Identify the function"]
    B --> C["Check the RANGE"]
    C --> D{"Argument positive<br/>or negative?"}
    D -->|"Positive"| E["Answer in first quadrant<br/>(straightforward)"]
    D -->|"Negative, sin⁻¹ or tan⁻¹"| F["Answer is NEGATIVE<br/>(range includes negatives)"]
    D -->|"Negative, cos⁻¹ or cot⁻¹"| G["Answer in second quadrant<br/>(range is [0, π])"]
    A --> H{"Key identities"}
    H --> I["sin⁻¹(-x) = -sin⁻¹(x)"]
    H --> J["cos⁻¹(-x) = π - cos⁻¹(x)"]
    H --> K["tan⁻¹(-x) = -tan⁻¹(x)"]
    H --> L["sin⁻¹(x) + cos⁻¹(x) = π/2"]

The identity $\sin^{-1}x + \cos^{-1}x = \pi/2$ is powerful — if you know one, you immediately know the other. For negative arguments: $\sin^{-1}$ and $\tan^{-1}$ give negative answers (odd functions), while $\cos^{-1}$ and $\cot^{-1}$ give answers in the second quadrant.

Alternative Method — Use Reference Angles

Ignore the negative sign and find the reference angle for the positive value
Then adjust based on the function and its range:
- $\sin^{-1}(-x) = -\sin^{-1}(x)$ → negate the answer
- $\cos^{-1}(-x) = \pi - \cos^{-1}(x)$ → subtract from $\pi$
- $\tan^{-1}(-x) = -\tan^{-1}(x)$ → negate the answer

For JEE: memorise $\sin^{-1}x + \cos^{-1}x = \pi/2$ and $\tan^{-1}x + \cot^{-1}x = \pi/2$ . These identities simplify many problems instantly. Also, $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x+y}{1-xy}$ (when $xy < 1$ ) is essential for JEE Main numericals.

Common Mistake

The most frequent error: writing $\sin^{-1}(-1/2) = 11\pi/6$ or $330°$ . While $\sin(330°) = -1/2$ is correct, $330°$ is NOT in the principal range $[-\pi/2, \pi/2]$ . The answer must be $-\pi/6$ (which is $-30°$ ). Always check that your answer falls within the specified range. CBSE board exams deduct full marks for answers outside the principal range.

Question

Solution — Step by Step

Why This Works

Alternative Method — Use Reference Angles

Common Mistake

Try These Next