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Inverse Trigonometric Functions - Definition, range, domain, principal value branch

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Function Invertibility and Restriction: Trigonometric functions are periodic and hence not one-to-one over their natural domains. To define their inverses, their domains are restricted to intervals where they are bijective. For instance, sin⁡x\sin x is restricted to [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}] to ensure every value in the range [−1,1][-1, 1] is mapped from exactly one value in the domain.

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Principal Value Branch: The specific range of an inverse trigonometric function that is standardly chosen is called the Principal Value Branch. For y=sin⁡−1xy = \sin^{-1} x, the principal value branch is [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]. Any value of the inverse function lying in this range is called the Principal Value.

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Graphical Representation: The graph of an inverse trigonometric function is obtained by reflecting the graph of the original restricted trigonometric function about the line y=xy = x. Visually, if the sine curve moves along the x-axis, the sin⁡−1x\sin^{-1} x curve 'climbs' along the y-axis, bounded horizontally between x=−1x = -1 and x=1x = 1.

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Domain and Range of Inverse Functions: The domain of an inverse function is the range of the original function. For y=sin⁡−1xy = \sin^{-1} x and y=cos⁡−1xy = \cos^{-1} x, the domain is [−1,1][-1, 1]. For y=tan⁡−1xy = \tan^{-1} x and y=cot⁡−1xy = \cot^{-1} x, the domain is the set of all real numbers R\mathbb{R}.

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Visual Asymptotes in Tangent and Cotangent: The graphs of y=tan⁡−1xy = \tan^{-1} x and y=cot⁡−1xy = \cot^{-1} x feature horizontal asymptotes. For tan⁡−1x\tan^{-1} x, the graph approaches but never touches the lines y=π2y = \frac{\pi}{2} and y=−π2y = -\frac{\pi}{2} as xx goes to ∞\infty and −∞-\infty respectively.

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Notational Distinction: It is critical to note that sin⁡−1x\sin^{-1} x is NOT equal to (sin⁡x)−1(\sin x)^{-1}. The former represents the inverse function (an angle), while the latter represents the reciprocal 1sin⁡x=csc⁡x\frac{1}{\sin x} = \csc x.

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Continuity and Monotonicity: Within their principal value branches, sin⁡−1x\sin^{-1} x, tan⁡−1x\tan^{-1} x, and csc⁡−1x\csc^{-1} x are increasing functions, whereas cos⁡−1x\cos^{-1} x, cot⁡−1x\cot^{-1} x, and sec⁡−1x\sec^{-1} x are decreasing functions.

📐Formulae

sin⁡−1:[−1,1]→[−π2,π2]\sin^{-1}: [-1, 1] \rightarrow [-\frac{\pi}{2}, \frac{\pi}{2}]

cos⁡−1:[−1,1]→[0,π]\cos^{-1}: [-1, 1] \rightarrow [0, \pi]

tan⁡−1:R→(−π2,π2)\tan^{-1}: \mathbb{R} \rightarrow (-\frac{\pi}{2}, \frac{\pi}{2})

cot⁡−1:R→(0,π)\cot^{-1}: \mathbb{R} \rightarrow (0, \pi)

sec⁡−1:R−(−1,1)→[0,π]−{π2}\sec^{-1}: \mathbb{R} - (-1, 1) \rightarrow [0, \pi] - \{\frac{\pi}{2}\}

csc⁡−1:R−(−1,1)→[−π2,π2]−{0}\csc^{-1}: \mathbb{R} - (-1, 1) \rightarrow [-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}

sin⁡−1(−x)=−sin⁡−1x,x∈[−1,1]\sin^{-1}(-x) = -\sin^{-1} x, x \in [-1, 1]

cos⁡−1(−x)=π−cos⁡−1x,x∈[−1,1]\cos^{-1}(-x) = \pi - \cos^{-1} x, x \in [-1, 1]

tan⁡−1(−x)=−tan⁡−1x,x∈R\tan^{-1}(-x) = -\tan^{-1} x, x \in \mathbb{R}

csc⁡−1(1x)=sin⁡−1x,∣x∣≤1,x≠0\csc^{-1}(\frac{1}{x}) = \sin^{-1} x, |x| \le 1, x \neq 0

💡Examples

Problem 1:

Find the principal value of cos⁡−1(−12)\cos^{-1}(-\frac{1}{2}).

Solution:

  1. Let y=cos⁡−1(−12)y = \cos^{-1}(-\frac{1}{2}). Then cos⁡y=−12\cos y = -\frac{1}{2}.
  2. We know that cos⁡(π3)=12\cos(\frac{\pi}{3}) = \frac{1}{2}.
  3. Since the range of the principal value branch of cos⁡−1\cos^{-1} is [0,π][0, \pi], we need a value in the second quadrant where cosine is negative.
  4. Using the identity cos⁡(π−θ)=−cos⁡θ\cos(\pi - \theta) = -\cos \theta, we get cos⁡(π−π3)=−cos⁡(π3)=−12\cos(\pi - \frac{\pi}{3}) = -\cos(\frac{\pi}{3}) = -\frac{1}{2}.
  5. Thus, cos⁡(2π3)=−12\cos(\frac{2\pi}{3}) = -\frac{1}{2}.
  6. Since 2π3∈[0,π]\frac{2\pi}{3} \in [0, \pi], the principal value is 2π3\frac{2\pi}{3}.

Explanation:

This problem uses the definition of the principal value branch for cosine, which requires the output angle to be between 00 and π\pi. Since the argument is negative, we use the second quadrant.

Problem 2:

Find the value of tan⁡−1(3)−sec⁡−1(−2)\tan^{-1}(\sqrt{3}) - \sec^{-1}(-2).

Solution:

  1. Let x=tan⁡−1(3)x = \tan^{-1}(\sqrt{3}). Since tan⁡(π3)=3\tan(\frac{\pi}{3}) = \sqrt{3} and π3∈(−π2,π2)\frac{\pi}{3} \in (-\frac{\pi}{2}, \frac{\pi}{2}), x=π3x = \frac{\pi}{3}.
  2. Let y=sec⁡−1(−2)y = \sec^{-1}(-2). Then sec⁡y=−2\sec y = -2, which means cos⁡y=−12\cos y = -\frac{1}{2}.
  3. From the principal branch of sec⁡−1\sec^{-1}, y∈[0,π]−{π2}y \in [0, \pi] - \{\frac{\pi}{2}\}.
  4. cos⁡y=−12  ⟹  y=π−π3=2π3\cos y = -\frac{1}{2} \implies y = \pi - \frac{\pi}{3} = \frac{2\pi}{3}.
  5. Now, tan⁡−1(3)−sec⁡−1(−2)=π3−2π3=−π3\tan^{-1}(\sqrt{3}) - \sec^{-1}(-2) = \frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3}.

Explanation:

To solve expressions with multiple inverse functions, calculate the principal value of each term individually according to their specific range restrictions, then perform the arithmetic.