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

medium CBSE JEE-MAIN • 3 min read
Tags Inverse Trig

Question

Find: (a) sin⁡−1(12)\sin^{-1}(\frac{1}{2}), (b) cos⁡−1(−32)\cos^{-1}(-\frac{\sqrt{3}}{2}), (c) tan⁡−1(−1)\tan^{-1}(-1). State the principal value branch for each inverse trig function.

Solution — Step by Step

FunctionDomainPrincipal Value Range
sin⁡−1x\sin^{-1}x[−1,1][-1, 1][−π/2,π/2][-\pi/2, \pi/2]
cos⁡−1x\cos^{-1}x[−1,1][-1, 1][0,π][0, \pi]
tan⁡−1x\tan^{-1}x(−∞,∞)(-\infty, \infty)(−π/2,π/2)(-\pi/2, \pi/2)

(a) sin⁡−1(1/2)\sin^{-1}(1/2): We need θ∈[−π/2,π/2]\theta \in [-\pi/2, \pi/2] such that sin⁡θ=1/2\sin\theta = 1/2. That is θ=π/6\theta = \mathbf{\pi/6} (or 30°).

(b) cos⁡−1(−3/2)\cos^{-1}(-\sqrt{3}/2): We need θ∈[0,π]\theta \in [0, \pi] such that cos⁡θ=−3/2\cos\theta = -\sqrt{3}/2. In the second quadrant, cos⁡(5π/6)=−3/2\cos(5\pi/6) = -\sqrt{3}/2. So θ=5π/6\theta = \mathbf{5\pi/6} (or 150°).

(c) tan⁡−1(−1)\tan^{-1}(-1): We need θ∈(−π/2,π/2)\theta \in (-\pi/2, \pi/2) such that tan⁡θ=−1\tan\theta = -1. That is θ=−π/4\theta = \mathbf{-\pi/4} (or −45°-45°).

Why This Works

graph TD
    A["Inverse Trig: Which quadrant?"] --> B["sin⁻¹: range -π/2 to π/2"]
    B --> C["Positive input → Q1 angle"]
    B --> D["Negative input → Q4 angle negative"]
    A --> E["cos⁻¹: range 0 to π"]
    E --> F["Positive input → Q1 angle"]
    E --> G["Negative input → Q2 angle"]
    A --> H["tan⁻¹: range -π/2 to π/2"]
    H --> I["Positive input → Q1 angle"]
    H --> J["Negative input → Q4 angle negative"]

The trig functions are not one-to-one over their full domain — sin⁡\sin has many angles giving the same value (e.g., sin⁡30°=sin⁡150°=1/2\sin 30° = \sin 150° = 1/2). To define an inverse, we restrict the domain to a branch where the function IS one-to-one. This restricted range is the principal value branch.

The choices are not arbitrary: they are selected to make the function continuous and to cover the full range of output values exactly once.

Alternative Method

Key identities for negative arguments:

  • sin⁡−1(−x)=−sin⁡−1(x)\sin^{-1}(-x) = -\sin^{-1}(x) (odd function)
  • cos⁡−1(−x)=π−cos⁡−1(x)\cos^{-1}(-x) = \pi - \cos^{-1}(x)
  • tan⁡−1(−x)=−tan⁡−1(x)\tan^{-1}(-x) = -\tan^{-1}(x) (odd function)

So for (b): cos⁡−1(−3/2)=π−cos⁡−1(3/2)=π−π/6=5π/6\cos^{-1}(-\sqrt{3}/2) = \pi - \cos^{-1}(\sqrt{3}/2) = \pi - \pi/6 = 5\pi/6.

These identities convert negative-argument problems into positive-argument ones, which are easier to evaluate.

Common Mistake

Writing sin⁡−1(1/2)=150°\sin^{-1}(1/2) = 150° or cos⁡−1(−3/2)=−30°\cos^{-1}(-\sqrt{3}/2) = -30°. These values are outside the principal value range. sin⁡−1\sin^{-1} always returns a value in [−π/2,π/2][-\pi/2, \pi/2], so it cannot return 150°. cos⁡−1\cos^{-1} always returns a value in [0,π][0, \pi], so it cannot return a negative angle. Always check that your answer falls within the principal branch before writing it as final.

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

tan⁡−1x+cot⁡−1x=π/2\tan^{-1}x + \cot^{-1}x = \pi/2 for all xx

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

2tan⁡−1x=sin⁡−12x1+x2=cos⁡−11−x21+x22\tan^{-1}x = \sin^{-1}\frac{2x}{1+x^2} = \cos^{-1}\frac{1-x^2}{1+x^2}

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