Inverse Trigonometric Functions: Numerical Problems Set (7)

easy • 2 min read
Tags Inverse Trigonometric Functions

Question

Evaluate sin⁡−1(12)+cos⁡−1(12)\sin^{-1}\left(\dfrac{1}{2}\right) + \cos^{-1}\left(\dfrac{1}{2}\right).

Solution — Step by Step

sin⁡−1:[−1,1]→[−π/2,π/2]\sin^{-1}: [-1, 1] \to [-\pi/2, \pi/2].

cos⁡−1:[−1,1]→[0,π]\cos^{-1}: [-1, 1] \to [0, \pi].

So both inverses give angles in their principal ranges.

sin⁡−1(1/2)=π/6\sin^{-1}(1/2) = \pi/6 (since sin⁡(π/6)=1/2\sin(\pi/6) = 1/2 and π/6∈[−π/2,π/2]\pi/6 \in [-\pi/2, \pi/2]).

cos⁡−1(1/2)=π/3\cos^{-1}(1/2) = \pi/3 (since cos⁡(π/3)=1/2\cos(\pi/3) = 1/2 and π/3∈[0,π]\pi/3 \in [0, \pi]).

sin⁡−1(1/2)+cos⁡−1(1/2)=π6+π3=π6+2π6=3π6=π2\sin^{-1}(1/2) + \cos^{-1}(1/2) = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{6} + \frac{2\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}

The identity sin⁡−1x+cos⁡−1x=π/2\sin^{-1}x + \cos^{-1}x = \pi/2 for x∈[−1,1]x \in [-1, 1] gives the answer in one step. ✓

Final answer: π2\dfrac{\pi}{2}.

Why This Works

The identity sin⁡−1x+cos⁡−1x=π/2\sin^{-1}x + \cos^{-1}x = \pi/2 comes from the fact that sin⁡θ=cos⁡(π/2−θ)\sin\theta = \cos(\pi/2 - \theta). If sin⁡−1x=α\sin^{-1}x = \alpha and cos⁡−1x=β\cos^{-1}x = \beta, then sin⁡α=x=cos⁡β=sin⁡(π/2−β)\sin\alpha = x = \cos\beta = \sin(\pi/2 - \beta), so α=π/2−β\alpha = \pi/2 - \beta, i.e., α+β=π/2\alpha + \beta = \pi/2.

This works for all x∈[−1,1]x \in [-1, 1] — no exceptions, no edge cases.

sin⁡−1x+cos⁡−1x=π2\sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2} for x∈[−1,1]x \in [-1, 1]

tan⁡−1x+cot⁡−1x=π2\tan^{-1}x + \cot^{-1}x = \dfrac{\pi}{2} for x∈Rx \in \mathbb{R}

sec⁡−1x+csc⁡−1x=π2\sec^{-1}x + \csc^{-1}x = \dfrac{\pi}{2} for ∣x∣≥1|x| \geq 1

tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy)\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\dfrac{x+y}{1-xy}\right) when xy<1xy < 1

(For xy>1xy > 1, add π\pi if both positive, subtract π\pi if both negative.)

Alternative Method

Direct computation as in steps 1-3, without using the identity. Both approaches give π/2\pi/2. Use the identity when you spot it — it saves time.

For CBSE 2-mark questions, the identity-based approach is slick. For 4-mark proofs, write out both inverse values explicitly to show working.

Common Mistake

Students write sin⁡−1(1/2)=30°\sin^{-1}(1/2) = 30° or sin⁡−1(1/2)=π/6\sin^{-1}(1/2) = \pi/6 rad — both are correct, but mixing units in one expression is a marks-loser. Stick to radians for inverse trig in calculus contexts.

Another classic: sin⁡−1(sin⁡x)=x\sin^{-1}(\sin x) = x. This is only true when x∈[−π/2,π/2]x \in [-\pi/2, \pi/2]. Outside this range, sin⁡−1(sin⁡x)≠x\sin^{-1}(\sin x) \neq x in general — it’s some reflected/wrapped version. JEE Main 2022 had a question targeting exactly this confusion.

A third trap: cos⁡−1(1/2)\cos^{-1}(1/2) is not −π/3-\pi/3 — even though cos⁡(−π/3)=1/2\cos(-\pi/3) = 1/2, the principal value range is [0,π][0, \pi], so the answer is +π/3+\pi/3 only.

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