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Inverse Trigonometric Functions - Basic Concepts

Grade 12CBSE

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

🔑Concepts

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The inverse sine function y=sin⁡−1xy = \sin^{-1} x is the inverse of the restricted sine function sin⁡:[−π2,π2]→[−1,1]\sin: [-\frac{\pi}{2}, \frac{\pi}{2}] \rightarrow [-1, 1]. Its domain is [−1,1][-1, 1] and its principal value branch (range) is [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]. The graph shows the reflection of sin⁡x\sin x across the line y=xy=x.

Graph of inverse sine function showing the reflection across y=x
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The principal value of an inverse trigonometric function is the value that lies in the defined range of the principal branch. For example, for cos⁡−1x\cos^{-1} x, the result must be in the interval [0,π][0, \pi].

Semicircle representing the principal value branch of cosine inverse from 0 to pi
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For the function y=tan⁡−1xy = \tan^{-1} x, the domain is the set of all real numbers R\mathbb{R}, and the range (principal branch) is the open interval (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}). Horizontal asymptotes occur at y=π2y = \frac{\pi}{2} and y=−π2y = -\frac{\pi}{2}.

Asymptotes for the arctan function at plus and minus pi/2
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Negative arguments in inverse functions follow specific rules: sin⁡−1(−x)=−sin⁡−1(x)\sin^{-1}(-x) = -\sin^{-1}(x) (odd symmetry), whereas cos⁡−1(−x)=π−cos⁡−1(x)\cos^{-1}(-x) = \pi - \cos^{-1}(x) due to the range being restricted to [0,π][0, \pi].

📐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})

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

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

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

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

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

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

💡Examples

Problem 1:

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

Solution:

Let y=sin⁡−1(12)y = \sin^{-1}\left(\frac{1}{2}\right). Then sin⁡y=12\sin y = \frac{1}{2}. We know that sin⁡(π6)=12\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}. Since π6∈[−π2,π2]\frac{\pi}{6} \in [-\frac{\pi}{2}, \frac{\pi}{2}], the principal value is π6\frac{\pi}{6}.

Explanation:

The value must fall within the principal value branch of sin⁡−1x\sin^{-1} x, which is [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}].

Problem 2:

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

Solution:

Let y=cos⁡−1(−12)y = \cos^{-1}\left(-\frac{1}{2}\right). Then cos⁡y=−12\cos y = -\frac{1}{2}. Since cos⁡(π3)=12\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}, we have cos⁡y=−cos⁡(π3)=cos⁡(π−π3)=cos⁡(2π3)\cos y = -\cos\left(\frac{\pi}{3}\right) = \cos\left(\pi - \frac{\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right). Since 2π3∈[0,π]\frac{2\pi}{3} \in [0, \pi], the principal value is 2π3\frac{2\pi}{3}.

Explanation:

For negative arguments in cos⁡−1\cos^{-1}, we use the property cos⁡−1(−x)=π−cos⁡−1x\cos^{-1}(-x) = \pi - \cos^{-1} x to ensure the result is within the range [0,π][0, \pi].

Problem 3:

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

Solution:

  1. tan⁡−1(3)=π3\tan^{-1}(\sqrt{3}) = \frac{\pi}{3} because tan⁡(π3)=3\tan\left(\frac{\pi}{3}\right) = \sqrt{3} and π3∈(−π2,π2)\frac{\pi}{3} \in (-\frac{\pi}{2}, \frac{\pi}{2}).
  2. sec⁡−1(−2)=π−sec⁡−1(2)\sec^{-1}(-2) = \pi - \sec^{-1}(2). Since sec⁡(π3)=2\sec\left(\frac{\pi}{3}\right) = 2, sec⁡−1(−2)=π−π3=2π3\sec^{-1}(-2) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}.
  3. Value =π3−2π3=−π3= \frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3}.

Explanation:

The problem is solved by calculating the principal values of each term individually and then performing subtraction.

Problem 4:

Evaluate the principal value of sin⁡−1(sin⁡2π3)\sin^{-1}\left(\sin \frac{2\pi}{3}\right).

Unit circle showing the relationship between 2pi/3 and pi/3

Solution:

  1. Note that 2π3\frac{2\pi}{3} does not lie in the principal branch [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}].
  2. We use the identity sin⁡θ=sin⁡(π−θ)\sin \theta = \sin(\pi - \theta).
  3. sin⁡2π3=sin⁡(π−2π3)=sin⁡π3\sin \frac{2\pi}{3} = \sin(\pi - \frac{2\pi}{3}) = \sin \frac{\pi}{3}.
  4. Now, π3∈[−π2,π2]\frac{\pi}{3} \in [-\frac{\pi}{2}, \frac{\pi}{2}].
  5. sin⁡−1(sin⁡π3)=π3\sin^{-1}\left(\sin \frac{\pi}{3}\right) = \frac{\pi}{3}.

Explanation:

Since the input to the inverse sine function must be within its principal range to 'cancel' the sine, we reduce the angle using trigonometric identities until it falls within [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}].

Problem 5:

Find the value of cos⁡−1(cos⁡7π6)\cos^{-1}\left(\cos \frac{7\pi}{6}\right).

Unit circle showing 7pi/6 in third quadrant and 5pi/6 in second quadrant

Solution:

  1. 7π6\frac{7\pi}{6} is outside the principal branch [0,π][0, \pi].
  2. Rewrite cos⁡7π6\cos \frac{7\pi}{6} using the property cos⁡(2π−θ)=cos⁡θ\cos(2\pi - \theta) = \cos \theta.
  3. cos⁡7π6=cos⁡(2π−7π6)=cos⁡5π6\cos \frac{7\pi}{6} = \cos(2\pi - \frac{7\pi}{6}) = \cos \frac{5\pi}{6}.
  4. 5π6\frac{5\pi}{6} lies in the interval [0,π][0, \pi].
  5. Therefore, cos⁡−1(cos⁡5π6)=5π6\cos^{-1}\left(\cos \frac{5\pi}{6}\right) = \frac{5\pi}{6}.

Explanation:

To find the principal value, the angle must be mapped to the interval [0,π][0, \pi] where the cosine function is one-to-one.