sin⁻¹(1/2) = π/6 because sin(π/6) = 1/2. Range of sin⁻¹.

Find sin⁻¹(1/2) — Basic Inverse Trig

Question

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

Solution — Step by Step

$\sin^{-1}\left(\dfrac{1}{2}\right)$ does NOT mean $\dfrac{1}{\sin(1/2)}$ . It asks: which angle has a sine of $\dfrac{1}{2}$ ? We’re looking for $\theta$ such that $\sin\theta = \dfrac{1}{2}$ .

The range of $\sin^{-1}$ is $\left[-\dfrac{\pi}{2},\ \dfrac{\pi}{2}\right]$ . This is the restriction that makes $\sin^{-1}$ a proper function — without it, infinitely many angles satisfy $\sin\theta = \dfrac{1}{2}$ . We only want the answer that falls in this window.

From our trigonometry tables:

\sin\frac{\pi}{6} = \frac{1}{2}

Since $\dfrac{\pi}{6} \in \left[-\dfrac{\pi}{2},\ \dfrac{\pi}{2}\right]$ , it qualifies as the principal value.

\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}

Answer: $\dfrac{\pi}{6}$ (or equivalently, 30°)

Why This Works

The function $\sin^{-1}$ (read: “arc sine”) is the inverse of $\sin$ restricted to $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ . That restriction is not arbitrary — it is the largest interval around 0 where $\sin$ is one-to-one, so its inverse exists cleanly.

When we write $y = \sin^{-1}(x)$ , we are saying $\sin y = x$ with $y$ forced into $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ . For $x = \dfrac{1}{2}$ , the unique $y$ in that range is $\dfrac{\pi}{6}$ .

This is why memorising the sine values of $0, \dfrac{\pi}{6}, \dfrac{\pi}{4}, \dfrac{\pi}{3}, \dfrac{\pi}{2}$ is non-negotiable in Class 12. Every inverse trig question at NCERT and CBSE board level comes back to this table.

Alternative Method

We can verify using the definition directly. If $\theta = \sin^{-1}\left(\dfrac{1}{2}\right)$ , then:

\sin\theta = \frac{1}{2} \quad \text{and} \quad \theta \in \left[-\frac{\pi}{2},\ \frac{\pi}{2}\right]

Draw a right triangle with opposite side = 1, hypotenuse = 2. By the Pythagorean theorem, adjacent side = $\sqrt{3}$ . This is the classic 30-60-90 triangle, confirming $\theta = 30° = \dfrac{\pi}{6}$ .

In CBSE board papers, answers in radians ( $\pi/6$ ) and degrees (30°) are both accepted unless the question specifies. JEE always expects radians. Get comfortable with both.

Common Mistake

Writing $\sin^{-1}\left(\dfrac{1}{2}\right) = \dfrac{5\pi}{6}$ .

Students remember that $\sin\left(\dfrac{5\pi}{6}\right) = \dfrac{1}{2}$ is also true — and it is. But $\dfrac{5\pi}{6}$ is outside the range $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ , so it is NOT a valid output of $\sin^{-1}$ . The whole point of the principal value range is to give a unique answer. Always check that your answer lies in $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ for $\sin^{-1}$ .

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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