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Class 12 Maths Notes

Inverse Trigonometric Functions Class 12 Notes

Exam-ready notes on inverse trigonometric functions: principal value branches, the restricted domains that make each trig function one-one, and the identities you need for board and JEE problems — with solved examples.

Class12SubjectMathematicsCoversCBSE · JEE

Written byDeep Narayan· Science & Mathematics EducatorReviewed byPushpanjali

What is this chapter about in one line?

Inverse trigonometric functions undo the six trig functions, but only after each function is restricted to a principal value branch where it is one-one.

Principal Value Branches

Principal value

Trigonometric functions are not one-one on their natural domain, so their inverses are defined on restricted intervals. The value chosen on that interval is the principal value. For example sin⁡−1x\sin^{-1} x is defined on [−1,1][-1, 1] with values in [−π/2, π/2][-\pi/2,\ \pi/2].

sin⁡−1:[−1,1]→[−π/2, π/2],cos⁡−1:[−1,1]→[0, π],tan⁡−1:R→(−π/2, π/2)\sin^{-1}: [-1,1] \to [-\pi/2,\ \pi/2],\quad \cos^{-1}: [-1,1] \to [0,\ \pi],\quad \tan^{-1}: \mathbb{R} \to (-\pi/2,\ \pi/2)
Principal value ranges

Standard Values and Symmetry

  • If
  • x∈[−1,1]x \in [-1,1]
  • ,
  • sin⁡−1(sin⁡y)=y\sin^{-1}(\sin y) = y
  • only when
  • y∈[−π/2, π/2]y \in [-\pi/2,\ \pi/2]
  • , not outside the branch.

Branch trap

The identity sin⁻¹(sin y) = y fails outside the principal range. Always check that the angle lies inside the branch before cancelling an inverse against a trig function.

Key Identities

sin⁡−1(−x)=−sin⁡−1x,cos⁡−1(−x)=π−cos⁡−1x,tan⁡−1(−x)=−tan⁡−1x\sin^{-1}(-x) = -\sin^{-1} x,\quad \cos^{-1}(-x) = \pi - \cos^{-1} x,\quad \tan^{-1}(-x) = -\tan^{-1} x
Negatives
sin⁡−1x+cos⁡−1x=π2,tan⁡−1x+cot⁡−1x=π2\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2},\qquad \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}
Complementary pairs
tan⁡−1x+tan⁡−1y={tan⁡−1x+y1−xy,xy<1π+tan⁡−1x+y1−xy,x,y>0, xy>1\tan^{-1} x + \tan^{-1} y = \begin{cases}\tan^{-1}\frac{x+y}{1-xy}, & xy < 1\\ \pi + \tan^{-1}\frac{x+y}{1-xy}, & x,y > 0,\ xy > 1\end{cases}
Sum of inverse tangents

Solving Equations

  • Convert the equation into a single inverse function, then apply the appropriate identity.
  • Check every candidate solution against the principal value range.
  • Use substitutions like
  • tan⁡−1x=θ\tan^{-1} x = \theta
  • to turn an inverse-trig equation into a plain trig equation.

Solved Examples

Example: Find the principal value of sin⁡−1(−12)\sin^{-1}\left(-\frac{1}{2}\right).

Solution: On the principal branch [−π/2, π/2][-\pi/2,\ \pi/2], sin⁡(−π/6)=−1/2\sin(-\pi/6) = -1/2. Using the negative identity, sin⁡−1(−1/2)=−sin⁡−1(1/2)=−π/6\sin^{-1}(-1/2) = -\sin^{-1}(1/2) = -\pi/6.

Example: Evaluate tan⁡−1(1/2)+tan⁡−1(1/3)\tan^{-1}(1/2) + \tan^{-1}(1/3).

Solution: With xy=1/6<1xy = 1/6 < 1, the sum identity gives tan⁡−11/2+1/31−1/6=tan⁡−15/65/6=tan⁡−11=π/4\tan^{-1}\frac{1/2+1/3}{1-1/6} = \tan^{-1}\frac{5/6}{5/6} = \tan^{-1} 1 = \pi/4.

Revision

Key formulas at a glance

Memorise these before attempting numericals — most exam questions hinge on one of them.

Complementary pairs

sin⁡−1x+cos⁡−1x=π/2\sin^{-1} x + \cos^{-1} x = \pi/2

Negative of sine-inverse

sin⁡−1(−x)=−sin⁡−1x\sin^{-1}(-x) = -\sin^{-1} x

Negative of cosine-inverse

cos⁡−1(−x)=π−cos⁡−1x\cos^{-1}(-x) = \pi - \cos^{-1} x

Sum of inverse tangents

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

Cosine-inverse triple angle

2cos⁡−1x=cos⁡−1(2x2−1),3cos⁡−1x=cos⁡−1(4x3−3x)2\cos^{-1} x = \cos^{-1}(2x^2 - 1),\quad 3\cos^{-1} x = \cos^{-1}(4x^3 - 3x)

Sine-inverse triple angle

3sin⁡−1x=sin⁡−1(3x−4x3)3\sin^{-1} x = \sin^{-1}(3x - 4x^3)

Exam tips

How this chapter is asked

Where this topic appears in CBSE, JEE Main and NEET papers.

  • The principal value always lies inside the restricted branch.
  • sin⁻¹(sin y) = y holds only on [-π/2, π/2].
  • Pair complementary inverses together: sin⁻¹x + cos⁻¹x = π/2.
  • For tan⁻¹x + tan⁻¹y, the constant added (0 or π) depends on xy and the signs of x, y.
  • Never take the inverse of an angle already outside the principal range.

FAQ

Common questions

What does principal value mean in inverse trigonometry?

A principal value is the selected output of an inverse trigonometric function when the input belongs to its restricted — or principal — range, e.g. sin⁻¹x is always taken in [-π/2, π/2].

Why are inverse trigonometric functions defined on restricted domains?

Trigonometric functions repeat, so they are not one-one on their full domain and no inverse exists. Restricting each to an interval where it is one-one lets a well-defined inverse be built.

What is the formula for tan⁻¹x + tan⁻¹y?

tan⁻¹x + tan⁻¹y = tan⁻¹((x + y)/(1 − xy)) when xy < 1; for xy > 1 with x, y > 0 an extra π is added to keep the result in the principal range.

How do I find the principal value of sin⁻¹(−1/2)?

Use the negative identity: sin⁻¹(−1/2) = −sin⁻¹(1/2) = −π/6, which lies in the principal range [-π/2, π/2].

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