Inverse Trigonometric Functions

Properties of Inverse Trigonometric Functions

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Properties of Inverse Trigonometric Functions

Properties of Inverse Trigonometric Functions

Hello JEE aspirants! Inverse trigonometric functions are crucial for solving many problems in trigonometry and calculus. Mastering their properties will give you a significant edge in JEE Main. Let's dive in!

Conceptual Explanation

Think of inverse trigonometric functions as "angle finders." For example, sin⁡−1x\sin^{-1}x gives you the angle whose sine is xx. Understanding this basic concept is key to grasping the properties.

These properties allow us to manipulate and simplify expressions involving inverse trigonometric functions, making them easier to work with. They are derived from the properties of trigonometric functions themselves.

1. Negative Argument Properties

These properties deal with inverse trigonometric functions of negative arguments.

Formula 1: sin⁡−1(−x)=−sin⁡−1x\sin^{-1}(-x) = -\sin^{-1}x

sin⁡−1(−x)=−sin⁡−1x\sin^{-1}(-x) = -\sin^{-1}x

Derivation/Explanation: Let sin⁡−1(−x)=θ\sin^{-1}(-x) = \theta. Then, −x=sin⁡θ-x = \sin\theta, which implies x=−sin⁡θ=sin⁡(−θ)x = -\sin\theta = \sin(-\theta). Therefore, sin⁡−1x=−θ=−sin⁡−1(−x)\sin^{-1}x = -\theta = -\sin^{-1}(-x), and thus, sin⁡−1(−x)=−sin⁡−1x\sin^{-1}(-x) = -\sin^{-1}x. The range of sin⁡−1x\sin^{-1}x is [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}], so this property holds.

Example: sin⁡−1(−12)=−sin⁡−1(12)=−π6\sin^{-1}(-\frac{1}{2}) = -\sin^{-1}(\frac{1}{2}) = -\frac{\pi}{6}

Formula 2: cos⁡−1(−x)=π−cos⁡−1x\cos^{-1}(-x) = \pi - \cos^{-1}x

cos⁡−1(−x)=π−cos⁡−1x\cos^{-1}(-x) = \pi - \cos^{-1}x

Derivation/Explanation: Let cos⁡−1(−x)=θ\cos^{-1}(-x) = \theta. Then, −x=cos⁡θ-x = \cos\theta, which implies x=−cos⁡θ=cos⁡(π−θ)x = -\cos\theta = \cos(\pi - \theta). Therefore, cos⁡−1x=π−θ=π−cos⁡−1(−x)\cos^{-1}x = \pi - \theta = \pi - \cos^{-1}(-x), and thus, cos⁡−1(−x)=π−cos⁡−1x\cos^{-1}(-x) = \pi - \cos^{-1}x. The range of cos⁡−1x\cos^{-1}x is [0,π][0, \pi], so π−θ\pi - \theta remains within the range.

Example: cos⁡−1(−12)=π−cos⁡−1(12)=π−π3=2π3\cos^{-1}(-\frac{1}{2}) = \pi - \cos^{-1}(\frac{1}{2}) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}

Formula 3: tan⁡−1(−x)=−tan⁡−1x\tan^{-1}(-x) = -\tan^{-1}x

tan⁡−1(−x)=−tan⁡−1x\tan^{-1}(-x) = -\tan^{-1}x

Derivation/Explanation: Let tan⁡−1(−x)=θ\tan^{-1}(-x) = \theta. Then, −x=tan⁡θ-x = \tan\theta, which implies x=−tan⁡θ=tan⁡(−θ)x = -\tan\theta = \tan(-\theta). Therefore, tan⁡−1x=−θ=−tan⁡−1(−x)\tan^{-1}x = -\theta = -\tan^{-1}(-x), and thus, tan⁡−1(−x)=−tan⁡−1x\tan^{-1}(-x) = -\tan^{-1}x. The range of tan⁡−1x\tan^{-1}x is (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}), so this property holds.

Example: tan⁡−1(−1)=−tan⁡−1(1)=−π4\tan^{-1}(-1) = -\tan^{-1}(1) = -\frac{\pi}{4}

2. Complementary Function Properties

These properties relate inverse trigonometric functions of complementary angles.

Formula 4: sin⁡−1x+cos⁡−1x=π2\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}

sin⁡−1x+cos⁡−1x=π2\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}

Derivation/Explanation: Let sin⁡−1x=θ\sin^{-1}x = \theta. Then, x=sin⁡θ=cos⁡(π2−θ)x = \sin\theta = \cos(\frac{\pi}{2} - \theta). Therefore, cos⁡−1x=π2−θ=π2−sin⁡−1x\cos^{-1}x = \frac{\pi}{2} - \theta = \frac{\pi}{2} - \sin^{-1}x, and thus, sin⁡−1x+cos⁡−1x=π2\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}.

Example: If x=12x = \frac{1}{\sqrt{2}}, sin⁡−1(12)+cos⁡−1(12)=π4+π4=π2\sin^{-1}(\frac{1}{\sqrt{2}}) + \cos^{-1}(\frac{1}{\sqrt{2}}) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}

Formula 5: tan⁡−1x+cot⁡−1x=π2\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}

tan⁡−1x+cot⁡−1x=π2\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}

Derivation/Explanation: Let tan⁡−1x=θ\tan^{-1}x = \theta. Then, x=tan⁡θ=cot⁡(π2−θ)x = \tan\theta = \cot(\frac{\pi}{2} - \theta). Therefore, cot⁡−1x=π2−θ=π2−tan⁡−1x\cot^{-1}x = \frac{\pi}{2} - \theta = \frac{\pi}{2} - \tan^{-1}x, and thus, tan⁡−1x+cot⁡−1x=π2\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}.

Example: If x=1x = 1, tan⁡−1(1)+cot⁡−1(1)=π4+π4=π2\tan^{-1}(1) + \cot^{-1}(1) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}

Formula 6: sec⁡−1x+csc⁡−1x=π2\sec^{-1}x + \csc^{-1}x = \frac{\pi}{2}

sec⁡−1x+csc⁡−1x=π2\sec^{-1}x + \csc^{-1}x = \frac{\pi}{2}

Derivation/Explanation: Let sec⁡−1x=θ\sec^{-1}x = \theta. Then, x=sec⁡θ=csc⁡(π2−θ)x = \sec\theta = \csc(\frac{\pi}{2} - \theta). Therefore, csc⁡−1x=π2−θ=π2−sec⁡−1x\csc^{-1}x = \frac{\pi}{2} - \theta = \frac{\pi}{2} - \sec^{-1}x, and thus, sec⁡−1x+csc⁡−1x=π2\sec^{-1}x + \csc^{-1}x = \frac{\pi}{2}.

Example: If x=2x = \sqrt{2}, sec⁡−1(2)+csc⁡−1(2)=π4+π4=π2\sec^{-1}(\sqrt{2}) + \csc^{-1}(\sqrt{2}) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}

3. Reciprocal Argument Properties

Relate inverse functions with reciprocal arguments using reciprocal trigonometric identities.

These properties are inherently linked to complementary function properties and often used together.

4. Composition Properties

Composition properties combine trigonometric functions and their inverses. A simple case is sin⁡(sin⁡−1x)=x\sin(\sin^{-1}x) = x, but these can be trickier when considering ranges and domains.

For example: sin⁡(sin⁡−1x)=x\sin(\sin^{-1}x) = x for −1≤x≤1-1 \le x \le 1

Also: sin⁡−1(sin⁡x)=x\sin^{-1}(\sin x) = x for −π2≤x≤π2-\frac{\pi}{2} \le x \le \frac{\pi}{2}

It's essential to keep the domain and range in mind when working with these compositions.

Tips for Solving Problems

  • Always check the domain and range of the inverse trigonometric functions.
  • Use substitutions to simplify complex expressions. For example, let θ=sin⁡−1x\theta = \sin^{-1}x.
  • Convert to trigonometric functions to solve equations. For example, if sin⁡−1x=θ\sin^{-1}x = \theta, then x=sin⁡θx = \sin\theta.
  • Look for patterns and apply the appropriate property.

Common Mistakes to Avoid

  • Forgetting the domain and range of inverse trigonometric functions.
  • Incorrectly applying the properties, especially when dealing with negative arguments.
  • Not checking the solution to see if it satisfies the original equation.

JEE-Specific Tricks

  • Recognize that sin⁡−1x\sin^{-1}x, cos⁡−1x\cos^{-1}x, and tan⁡−1x\tan^{-1}x can be represented geometrically as angles in a right-angled triangle. Draw the triangle to visualize relationships.
  • Be prepared to solve problems that combine these properties with other concepts in trigonometry and calculus. JEE problems are often multi-concept.
  • Practice a variety of problems to build your problem-solving skills.
  • Use the complementary function properties to convert between sin⁡−1x\sin^{-1}x and cos⁡−1x\cos^{-1}x, tan⁡−1x\tan^{-1}x and cot⁡−1x\cot^{-1}x, etc.

That wraps up our lesson on properties of inverse trigonometric functions. Keep practicing, and you'll ace those JEE problems!