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Inverse Trigonometric Functions - Graphs of inverse trigonometric functions

Grade 12CBSE

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

🔑Concepts

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The graph of y=sin⁡−1xy = \sin^{-1} x is obtained by reflecting the graph of the sine function (restricted to its principal value branch [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]) about the line y=xy = x. Its domain is [−1,1][-1, 1] and range is [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}].

Symmetry of inverse sine function about y=x line.
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The graph of y=cos⁡−1xy = \cos^{-1} x has a domain of [−1,1][-1, 1] and a range (principal value branch) of [0,π][0, \pi]. It is a strictly decreasing function.

Graph of arccos(x) showing domain [-1, 1] and range [0, pi].
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The graph of y=tan⁡−1xy = \tan^{-1} x is continuous over the entire set of real numbers R\mathbb{R} and is bounded by horizontal asymptotes at y=π2y = \frac{\pi}{2} and y=−π2y = -\frac{\pi}{2}.

Graph of arctan(x) with horizontal asymptotes.
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For cosec−1x\text{cosec}^{-1} x and sec⁡−1x\sec^{-1} x, the graphs exist only for ∣x∣≥1|x| \geq 1. There is a 'gap' in the domain between −1-1 and 11 where the functions are not defined.

📐Formulae

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

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

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

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

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

cosec−1:R−(−1,1)→[−π2,π2]−{0}\text{cosec}^{-1}: \mathbb{R} - (-1, 1) \to [-\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}

💡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}).
  2. This implies cos⁡y=−12\cos y = -\frac{1}{2}.
  3. We know that the principal value branch of cos⁡−1\cos^{-1} is [0,π][0, \pi].
  4. Since cos⁡π3=12\cos \frac{\pi}{3} = \frac{1}{2}, we use the identity cos⁡(π−θ)=−cos⁡θ\cos(\pi - \theta) = -\cos \theta.
  5. cos⁡y=−cos⁡π3=cos⁡(π−π3)=cos⁡2π3\cos y = -\cos \frac{\pi}{3} = \cos(\pi - \frac{\pi}{3}) = \cos \frac{2\pi}{3}.
  6. Since 2π3∈[0,π]\frac{2\pi}{3} \in [0, \pi], the principal value is 2π3\frac{2\pi}{3}.

Explanation:

To find the principal value, we identify the angle in the restricted range [0,π][0, \pi] whose cosine equals the given value. Because the value is negative, the angle must lie in the second quadrant.

Problem 2:

Find the domain of the function f(x)=sin⁡−1(2x−3)f(x) = \sin^{-1}(2x - 3).

Solution:

  1. The domain of the basic function y=sin⁡−1ty = \sin^{-1} t is −1≤t≤1-1 \leq t \leq 1.
  2. For f(x)=sin⁡−1(2x−3)f(x) = \sin^{-1}(2x - 3), the argument (2x−3)(2x - 3) must lie within this range.
  3. Set up the inequality: −1≤2x−3≤1-1 \leq 2x - 3 \leq 1.
  4. Add 33 to all parts: 2≤2x≤42 \leq 2x \leq 4.
  5. Divide by 22: 1≤x≤21 \leq x \leq 2.
  6. Therefore, the domain is x∈[1,2]x \in [1, 2].

Explanation:

The domain of an inverse sine function is determined by ensuring the expression inside the function stays between −1-1 and 11, inclusive.

Problem 3:

Identify the quadrant and sign for the principal value of sin⁡−1(12)\sin^{-1}(\frac{1}{2}) and represent the angle on the unit circle.

Unit circle showing the angle 30 degrees in the first quadrant.

Solution:

Let θ=sin⁡−1(12)\theta = \sin^{-1}(\frac{1}{2}). Then sin⁡θ=12\sin \theta = \frac{1}{2}. Since the range of sin⁡−1\sin^{-1} is [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}] and the value is positive, θ\theta must lie in the First Quadrant. θ=π6\theta = \frac{\pi}{6} (or 30∘30^{\circ}).

Explanation:

The principal value of sin⁡−1x\sin^{-1} x for x>0x > 0 always falls in the first quadrant (0,π2](0, \frac{\pi}{2}].

Problem 4:

Sketch the region of the domain for y=sec⁡−1xy = \sec^{-1} x and determine if x=0.5x = 0.5 is a valid input.

Number line showing the domain of arcsec(x) with the central portion excluded.

Solution:

The domain of sec⁡−1x\sec^{-1} x is (−∞,−1]∪[1,∞)(-\infty, -1] \cup [1, \infty). Mathematically, ∣x∣≥1|x| \geq 1. Since 0.50.5 lies in the interval (−1,1)(-1, 1), it is not in the domain. Therefore, sec⁡−1(0.5)\sec^{-1}(0.5) is undefined.

Explanation:

Inverse secant is the inverse of the restricted secant function. Since ∣sec⁡θ∣≥1|\sec \theta| \geq 1, the domain of the inverse must reflect this range.