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02Class 12 Maths

Inverse Trigonometric Functions formulas

Domains, ranges and identities for Inverse Trigonometric Functions. Checking the principal value range is the whole trick of this chapter.

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Principal value ranges

sin⁻¹ x
x∈[−1,1],y∈[−π2,π2]\displaystyle x \in [-1, 1], \quad y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
cos⁻¹ x
x∈[−1,1],y∈[0,π]\displaystyle x \in [-1, 1], \quad y \in [0, \pi]
tan⁻¹ x
x∈R,y∈(−π2,π2)\displaystyle x \in \mathbb{R}, \quad y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

Identities

Negative arguments
sin⁡−1(−x)=−sin⁡−1x,tan⁡−1(−x)=−tan⁡−1x\displaystyle \sin^{-1}(-x) = -\sin^{-1}x, \quad \tan^{-1}(-x) = -\tan^{-1}x
For cosine
cos⁡−1(−x)=π−cos⁡−1x\displaystyle \cos^{-1}(-x) = \pi - \cos^{-1}x
Complementary pairs
sin⁡−1x+cos⁡−1x=π2,tan⁡−1x+cot⁡−1x=π2\displaystyle \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}, \quad \tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}
Sum of inverse tangents
tan⁡−1x+tan⁡−1y=tan⁡−1x+y1−xy\displaystyle \tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x + y}{1 - xy}

For xy < 1.

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

For suitable ranges of x.

Where students slip up

  • Assuming sin⁻¹(sin x) = x for any x. It only holds within the principal range.
  • Reading sin⁻¹ x as 1/sin x. It's the inverse function and not the reciprocal.

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