Problem 4 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Let \(f(x)=\sin ^{-1} x+\cos ^{-1} x\). Then \(\frac{\pi}{2}\) is equal to (a) \(f\left(\frac{1}{2}\right)\) (b) \(f\left(k^{2}-2 k+3\right), k \varepsilon R\) (c) \(f\left(\frac{1}{1+k^{2}}\right), k \varepsilon R\) (d) \(f(-2)\)

Step-by-Step Solution

Answer

The correct answers are options (a) and (c) as they fall within the defined range, hence, fulfilling the property \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \).

Step 1: Recognize Property of Inverse Trigonometric Functions

Firstly, it's important to understand a property of inverse trigonometric functions: \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \) for all \( x \) in the range -1 ≤ x ≤ 1. This is because the arc sine and arc cosine of any angle are supplementary.

Step 2: Check Option (a)

For option (a), substitute \( \frac{1}{2} \) into the function \( f(x) \). If this value lies in the defined range (-1 to 1), the function will yield \( \frac{\pi}{2} \). Since \( \frac{1}{2} \) is within the defined range, this option is valid.

Step 3: Check Option (b)

For option (b), substitute \( k^{2}-2k+3 \) into the function \( f(x) \). For all values of \( k \), \( k^{2}-2k+3 \) is greater than 1, thus it falls outside the defined range (-1 to 1). Therefore, this option is invalid.

Step 4: Check Option (c)

For option (c), substitute \( \frac{1}{1+k^{2}} \) into the function \( f(x) \). It can be seen that this expression will always be within the defined range (-1 to 1) for all real values of \( k \). Therefore, this option is valid.

Step 5: Check Option (d)

For option (d), substitute -2 into the function \( f(x) \). It falls outside the defined range (-1 to 1). Therefore, this option is invalid.

Key Concepts

Arc Sine and Arc Cosine Relationship

The functions arc sine, denoted as \( \sin^{-1} x \), and arc cosine, denoted as \( \cos^{-1} x \), have a special relationship. They are both inverse trigonometric functions and add up to \( \frac{\pi}{2} \) when their arguments are within a specific interval. This is because they represent complementary angles in the context of the unit circle.

In mathematical terms, for every \( x \) in the range \(-1 \leq x \leq 1\), the equation \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \) holds true. This reflects the relationship between sine and cosine where one function increases while the other decreases as the angle changes. Understanding this property helps in solving trigonometric problems easily.

Range of Inverse Trigonometric Functions

Inverse trigonometric functions have specific ranges in which they operate. This is crucial because for each value of \( x \), the inverse function returns a unique angle.

- Arc sine \( (\sin^{-1} x) \) has a range of \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).
- Arc cosine \( (\cos^{-1} x) \) typically ranges from \( [0, \pi] \).

These ranges are defined based on the principal values that make the inverse function a true inverse relation.

It is important to remember these ranges when solving trigonometric equations, as inputs falling outside the critical range \([-1, 1]\) do not produce valid angles in standard real number calculations.

Function Properties

Understanding the properties of inverse trigonometric functions can be a game-changer in solving related problems.

- **Domain:** For \( \sin^{-1} x \) and \( \cos^{-1} x \), the domain is \([-1, 1]\). Any value outside this range is not valid for real number solutions.
- **Adding Property:** The addition property \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \) simplifies many trigonometric equations and is particularly useful in solving problems efficiently.

These properties are foundational and are applied in understanding and solving various complex trigonometric problems systematically and accurately.

JEE Trigonometry Questions

The Joint Entrance Examination (JEE) often challenges students with questions involving inverse trigonometric functions. These questions often test understanding in creative ways.

- **Real-world applications:** Students may encounter problems that require application of properties such as \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \).
- **Domain and range validation:** It's common for questions to include conditions ensuring inputs are within valid domains, which requires checking the range of expressions like \( k^2 - 2k + 3 \).

To excel, it's important to understand fundamental principles and practice a variety of problems to get comfortable with the logic and mechanics of these functions. This insight aids in tackling similar challenges in JEE and other competitive exams effectively.