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Graphs and properties of inverse trig functions

Each inverse trigonometric function inherits properties from its parent. Monotonicity flips orientation only when the parent is decreasing, and the graph is always a reflection of the restricted parent across the line y=xy = x. This subtopic catalogues the six graphs and the symmetries you should know on sight.

The six graphs

y=sin⁡−1xy = \sin^{-1} x. Domain [−1,1][-1, 1], range [−π/2,π/2][-\pi/2, \pi/2]. Increasing. Passes through (−1,−π/2),(0,0),(1,π/2)(-1, -\pi/2), (0, 0), (1, \pi/2). Odd function: sin⁡−1(−x)=−sin⁡−1x\sin^{-1}(-x) = -\sin^{-1} x.

y=cos⁡−1xy = \cos^{-1} x. Domain [−1,1][-1, 1], range [0,π][0, \pi]. Decreasing. Passes through (−1,π),(0,π/2),(1,0)(-1, \pi), (0, \pi/2), (1, 0). Reflection of sin⁡−1\sin^{-1} about the line y=π/4y = \pi/4: in fact cos⁡−1x=π/2−sin⁡−1x\cos^{-1} x = \pi/2 - \sin^{-1} x.

y=tan⁡−1xy = \tan^{-1} x. Domain R\mathbb{R}, range (−π/2,π/2)(-\pi/2, \pi/2). Increasing. Horizontal asymptotes y=±π/2y = \pm \pi/2. Odd: tan⁡−1(−x)=−tan⁡−1x\tan^{-1}(-x) = -\tan^{-1} x.

y=cot⁡−1xy = \cot^{-1} x. Domain R\mathbb{R}, range (0,π)(0, \pi). Decreasing. Asymptotes y=0y = 0 (as x→∞x \to \infty) and y=πy = \pi (as x→−∞x \to -\infty). cot⁡−1(−x)=π−cot⁡−1x\cot^{-1}(-x) = \pi - \cot^{-1} x.

y=sec⁡−1xy = \sec^{-1} x. Domain ∣x∣≥1|x| \ge 1, range [0,π]∖{π/2}[0, \pi] \setminus \{\pi/2\}. Two branches: one on [1,∞)[1, \infty) increasing from 00 to π/2\pi/2, one on (−∞,−1](-\infty, -1] increasing from π/2\pi/2 to π\pi.

y=csc⁡−1xy = \csc^{-1} x. Domain ∣x∣≥1|x| \ge 1, range [−π/2,π/2]∖{0}[-\pi/2, \pi/2] \setminus \{0\}. Decreasing on each branch, symmetric to sec⁡−1\sec^{-1}.

Symmetry identities

IdentityReason
sin⁡−1(−x)=−sin⁡−1x\sin^{-1}(-x) = -\sin^{-1} xsin⁡\sin is odd on its branch
cos⁡−1(−x)=π−cos⁡−1x\cos^{-1}(-x) = \pi - \cos^{-1} xcos⁡\cos is even but the branch shifts the angle
tan⁡−1(−x)=−tan⁡−1x\tan^{-1}(-x) = -\tan^{-1} xtan⁡\tan is odd on its branch
cot⁡−1(−x)=π−cot⁡−1x\cot^{-1}(-x) = \pi - \cot^{-1} xcot⁡\cot shifts to the other branch

Monotonicity by inheritance

Since sin⁡\sin is increasing on [−π/2,π/2][-\pi/2, \pi/2], sin⁡−1\sin^{-1} is increasing on [−1,1][-1, 1]. Since cos⁡\cos is decreasing on [0,π][0, \pi], cos⁡−1\cos^{-1} is decreasing on [−1,1][-1, 1]. The same logic applies to the other four.

Derivatives (preview)

The derivatives are computed in Chapter 5, but it is useful to know:

FunctionDerivative
sin⁡−1x\sin^{-1} x11−x2\dfrac{1}{\sqrt{1 - x^2}}
cos⁡−1x\cos^{-1} x−11−x2-\dfrac{1}{\sqrt{1 - x^2}}
tan⁡−1x\tan^{-1} x11+x2\dfrac{1}{1 + x^2}
cot⁡−1x\cot^{-1} x−11+x2-\dfrac{1}{1 + x^2}
sec⁡−1x\sec^{-1} x1∥x∥x2−1\dfrac{1}{\|x\| \sqrt{x^2 - 1}}
csc⁡−1x\csc^{-1} x−1∥x∥x2−1-\dfrac{1}{\|x\| \sqrt{x^2 - 1}}

Reading the graph: an example

Suppose you are asked: for which xx is sin⁡−1x=tan⁡−1x\sin^{-1} x = \tan^{-1} x? From the graphs, both start at 00 for x=0x = 0 and both are increasing. sin⁡−1\sin^{-1} rises faster (its derivative 1/1−x21/\sqrt{1 - x^2} exceeds 1/(1+x2)1/(1 + x^2) for x∈(0,1)x \in (0, 1)). So the only intersection is at x=0x = 0.

Worked examples

Example 1. Without computing, decide whether sin⁡−1(0.6)>sin⁡−1(0.5)\sin^{-1}(0.6) > \sin^{-1}(0.5).

Yes, because sin⁡−1\sin^{-1} is strictly increasing on [−1,1][-1, 1].

Example 2. Decide whether cos⁡−1(0.4)>cos⁡−1(0.5)\cos^{-1}(0.4) > \cos^{-1}(0.5).

Yes, because cos⁡−1\cos^{-1} is strictly decreasing.

Example 3. Sketch y=sin⁡−1(sin⁡x)y = \sin^{-1}(\sin x) for x∈[−2π,2π]x \in [-2\pi, 2\pi].

Inside [−π/2,π/2][-\pi/2, \pi/2], the function is just xx. Outside, fold using sin⁡(π−x)=sin⁡x\sin(\pi - x) = \sin x. The graph is a triangular wave with peaks at ±π/2\pm \pi/2 and slope alternating between +1+1 and −1-1.

Example 4. Sketch y=cos⁡−1(cos⁡x)y = \cos^{-1}(\cos x) on [−2π,2π][-2\pi, 2\pi].

Inside [0,π][0, \pi], it is xx. Use cos⁡(−x)=cos⁡x\cos(-x) = \cos x and periodicity to fold. The graph is a triangular wave with peaks at π\pi and troughs at 00, 2π2\pi.

Example 5. Show that sin⁡−1x+cos⁡−1x=π/2\sin^{-1} x + \cos^{-1} x = \pi/2 for all x∈[−1,1]x \in [-1, 1].

Let θ=sin⁡−1x\theta = \sin^{-1} x, so x=sin⁡θx = \sin \theta and θ∈[−π/2,π/2]\theta \in [-\pi/2, \pi/2]. Then cos⁡(π/2−θ)=sin⁡θ=x\cos(\pi/2 - \theta) = \sin \theta = x, and π/2−θ∈[0,π]\pi/2 - \theta \in [0, \pi]. So cos⁡−1x=π/2−θ\cos^{-1} x = \pi/2 - \theta, giving the identity.

Example 6. Plot tan⁡−1x\tan^{-1} x and cot⁡−1x\cot^{-1} x on the same axes. Confirm that tan⁡−1x+cot⁡−1x=π/2\tan^{-1} x + \cot^{-1} x = \pi/2 for all xx.

The two curves are reflections of each other across y=π/4y = \pi/4, summing to a constant π/2\pi/2 at every xx.

Try it yourself

  1. Sketch y=sin⁡−1xy = \sin^{-1} x and mark its monotonicity.
  2. Find cos⁡−1(cos⁡(2.5))\cos^{-1}(\cos(2.5)) given π≈3.14\pi \approx 3.14.
  3. Show sin⁡−1(−x)=−sin⁡−1x\sin^{-1}(-x) = -\sin^{-1} x using the principal branch.
  4. Why is the graph of cot⁡−1x\cot^{-1} x continuous at x=0x = 0 but not differentiable in elementary ways?
  5. Sketch y=sin⁡−1x+cos⁡−1xy = \sin^{-1} x + \cos^{-1} x for x∈[−1,1]x \in [-1, 1] , note the constant value.
  6. Compute sec⁡−1(2)\sec^{-1}(\sqrt 2).
  7. Compute sec⁡−1(−2)\sec^{-1}(-\sqrt 2).
  8. Find the points where y=sin⁡−1(sin⁡x)y = \sin^{-1}(\sin x) has its peaks for x∈[−3π,3π]x \in [-3\pi, 3\pi].
  9. Sketch y=tan⁡−1(tan⁡x)y = \tan^{-1}(\tan x) for x∈[−π,π]x \in [-\pi, \pi].
  10. State the horizontal asymptotes of tan⁡−1x\tan^{-1} x.
  11. Why does cos⁡−1\cos^{-1} have no points of inflection?
  12. Verify the identity tan⁡−1x+cot⁡−1x=π/2\tan^{-1} x + \cot^{-1} x = \pi/2 by differentiating both sides.
  13. For which xx is sin⁡−1x=cos⁡−1x\sin^{-1} x = \cos^{-1} x?
  14. Show that sin⁡−1x+sin⁡−1(−x)=0\sin^{-1} x + \sin^{-1}(-x) = 0 for x∈[−1,1]x \in [-1, 1].

Pitfalls / Tricks

  • cos⁡−1\cos^{-1} is not odd. cos⁡−1(−x)=π−cos⁡−1x\cos^{-1}(-x) = \pi - \cos^{-1} x, not −cos⁡−1x-\cos^{-1} x.
  • The graph of sin⁡−1(sin⁡x)\sin^{-1}(\sin x) is piecewise linear , a sawtooth, not a smooth sine.
  • Conventions for sec⁡−1\sec^{-1} vary; in some books the range includes [π/2,π)[\pi/2, \pi) on the left branch instead of (π/2,π](\pi/2, \pi].
  • The derivative of sin⁡−1x\sin^{-1} x blows up at x=±1x = \pm 1 , the graph has vertical tangent at the endpoints.

Next we collect the standard inverse trigonometric identities.

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