Comprehensive Study Notes on Inverse Trigonometric Functions

Inverse Trigonometric Functions: Domain, Range, and Principal Value Branches

  • Definition and Scope: Inverse trigonometric functions are the inverse functions of the trigonometric functions with appropriately restricted domains. These are essential for solving trigonometric equations and are widely used in calculus and geometry.

  • Domain and Range (Principal Value Branch) Table: To ensure trigonometric functions are bijective (one-to-one and onto) and thus invertible, their domains must be restricted. The following table captures the standard domain and range (Principal Value Branch) for each function:

    • Function: sin⁡−1(x)\sin^{-1}(x)
      • Domain: [−1,1][-1, 1]
      • Principal Value Branch (Range): [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]
    • Function: cos⁡−1(x)\cos^{-1}(x)
      • Domain: [−1,1][-1, 1]
      • Principal Value Branch (Range): [0,π][0, \pi]
    • Function: tan⁡−1(x)\tan^{-1}(x)
      • Domain: R\mathbb{R} (all real numbers)
      • Principal Value Branch (Range): (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2})
    • Function: cot⁡−1(x)\cot^{-1}(x)
      • Domain: R\mathbb{R}
      • Principal Value Branch (Range): (0,π)(0, \pi)
    • Function: sec⁡−1(x)\sec^{-1}(x)
      • Domain: R−(−1,1)\mathbb{R} - (-1, 1) or (−∞,−1]∪[1,∞)(-\infty, -1] \cup [1, \infty)
      • Principal Value Branch (Range): [0,π]−{π2}[0, \pi] - \{\frac{\pi}{2}\}
    • Function: csc⁡−1(x)\csc^{-1}(x)
      • Domain: R−(−1,1)\mathbb{R} - (-1, 1) or (−∞,−1]∪[1,∞)(-\infty, -1] \cup [1, \infty)
      • Principal Value Branch (Range): [−π2,π2]−{0}[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}

Technical Simplifications and Evaluations (2-Mark Problems)

  • Evaluation of Mixed Inverse Expressions: Calculating the sum of specific inverse trigonometric values requires determining the angle within the principal value branch for each term.

    • Example: Find the value of cos⁡−1(12)+2sin⁡−1(12)\cos^{-1}(\frac{1}{2}) + 2\sin^{-1}(\frac{1}{2}).
      • Step 1: Evaluate cos⁡−1(12)\cos^{-1}(\frac{1}{2}). Since cos⁡(π3)=12\cos(\frac{\pi}{3}) = \frac{1}{2} and π3∈[0,π]\frac{\pi}{3} \in [0, \pi], the value is π3\frac{\pi}{3}.
      • Step 2: Evaluate sin⁡−1(12)\sin^{-1}(\frac{1}{2}). Since sin⁡(π6)=12\sin(\frac{\pi}{6}) = \frac{1}{2} and π6∈[−π2,π2]\frac{\pi}{6} \in [-\frac{\pi}{2}, \frac{\pi}{2}], the value is π6\frac{\pi}{6}.
      • Step 3: Combine the results: π3+2(π6)=π3+π3=2π3\frac{\pi}{3} + 2(\frac{\pi}{6}) = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3}.

  • Transformation Identities involving Radicals:

    • Identity for Sine Subsitution: sin⁡−1(2x1−x2)=2sin⁡−1(x)\sin^{-1}(2x\sqrt{1-x^2}) = 2\sin^{-1}(x).
      • This identity is derived using the substitution x=sin⁡(θ)x = \sin(\theta), leading to sin⁡−1(2sin⁡(θ)cos⁡(θ))=sin⁡−1(sin⁡(2θ))=2θ\sin^{-1}(2\sin(\theta)\cos(\theta)) = \sin^{-1}(\sin(2\theta)) = 2\theta.
    • Identity for Cosine Substitution: sin⁡−1(2x1−x2)=2cos⁡−1(x)\sin^{-1}(2x\sqrt{1-x^2}) = 2\cos^{-1}(x).
      • This identity is used when the domain of xx is restricted such that substitution via x=cos⁡(θ)x = \cos(\theta) is more appropriate.

  • Multiple Angle Formulas for Inverse Functions:

    • Triple Angle Sine Identity: 3sin⁡−1(x)=sin⁡−1(3x−4x3)3\sin^{-1}(x) = \sin^{-1}(3x - 4x^3).
      • This follows from the trigonometric identity sin⁡(3θ)=3sin⁡(θ)−4sin⁡3(θ)\sin(3\theta) = 3\sin(\theta) - 4\sin^3(\theta).
    • Triple Angle Cosine Identity: 3cos⁡−1(x)=cos⁡−1(4x3−3x)3\cos^{-1}(x) = \cos^{-1}(4x^3 - 3x).
      • This follows from the trigonometric identity cos⁡(3θ)=4cos⁡3(θ)−3cos⁡(θ)\cos(3\theta) = 4\cos^3(\theta) - 3\cos(\theta).

  • Simplification of Trigonometric Fractions:

    • Expression involving cot-1: Simplification of cot⁡−1(1−cos⁡(x)1+cos⁡(x))\cot^{-1}\left(\sqrt{\frac{1-\cos(x)}{1+\cos(x)}}\right).
      • Using half-angle formulas: 1−cos⁡(x)=2sin⁡2(x2)1-\cos(x) = 2\sin^2(\frac{x}{2}) and 1+cos⁡(x)=2cos⁡2(x2)1+\cos(x) = 2\cos^2(\frac{x}{2}).
      • The expression simplifies to cot⁡−1(2sin⁡2(x2)2cos⁡2(x2))=cot⁡−1(tan⁡2(x2))=cot⁡−1(tan⁡(x2))\cot^{-1}\left(\sqrt{\frac{2\sin^2(\frac{x}{2})}{2\cos^2(\frac{x}{2})}}\right) = \cot^{-1}\left(\sqrt{\tan^2(\frac{x}{2})}\right) = \cot^{-1}(\tan(\frac{x}{2}))
      • Further reduction: cot⁡−1(cot⁡(π2−x2))=π2−x2\cot^{-1}(\cot(\frac{\pi}{2} - \frac{x}{2})) = \frac{\pi}{2} - \frac{x}{2}.
    • Expression involving tan-1: Simplification of tan⁡−1(cos⁡(x)1−sin⁡(x))\tan^{-1}\left(\frac{\cos(x)}{1-\sin(x)}\right).
      • Using identities: cos⁡(x)=sin⁡(π2−x)\cos(x) = \sin(\frac{\pi}{2} - x) and 1−sin⁡(x)=1−cos⁡(π2−x)1-\sin(x) = 1-\cos(\frac{\pi}{2} - x).

  • Compound and Compositional Evaluations:

    • Nested Evaluation: Evaluate tan⁡−1[2cos⁡(2sin⁡−1(12))]\tan^{-1}\left[2\cos(2\sin^{-1}(\frac{1}{2}))\right].
      • Step 1: Inside-out evaluation starts with sin⁡−1(12)=π6\sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}.
      • Step 2: Multiply by 2: 2×π6=π32 \times \frac{\pi}{6} = \frac{\pi}{3}.
      • Step 3: Evaluate cos⁡(π3)=12\cos(\frac{\pi}{3}) = \frac{1}{2}.
      • Step 4: Multiply by 2: 2×12=12 \times \frac{1}{2} = 1.
      • Step 5: Final calculation: tan⁡−1(1)=π4\tan^{-1}(1) = \frac{\pi}{4}.
    • Principal Value beyond standard cycles: Find the value of cos⁡−1(cos⁡(13π6))\cos^{-1}\left(\cos(\frac{13\pi}{6})\right).
      • Note that 13π6\frac{13\pi}{6} is not in the principal range [0,π][0, \pi].
      • Using the periodicity of cosine: cos⁡(13π6)=cos⁡(2π+π6)=cos⁡(π6)\cos(\frac{13\pi}{6}) = \cos(2\pi + \frac{\pi}{6}) = \cos(\frac{\pi}{6}).
      • Since π6∈[0,π]\frac{\pi}{6} \in [0, \pi], the final result is π6\frac{\pi}{6}.