Problem 41 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Solve the following equations and tick the correct one. If \(\sin x+\cos x=\sqrt{y+\frac{1}{y}}, x \in[0, \pi]\), then (a) \(x=\frac{\pi}{4}, y=1\) (b) \(y=0\) (c) \(y=2\) (d) \(x=\frac{3 \pi}{4}\)

Step-by-Step Solution

Answer

All the given options (a), (b), (c) and (d) are incorrect based on the provided equation and the given range of variable x.

Step 1 Evaluating option (a)

Here, x = \(\frac{\pi}{4}\) and y = 1. Substituting these values into the equation, we get \(\sin(\frac{\pi}{4}) + \cos(\frac{\pi}{4}) = \sqrt{1+1}\). Solving this, we get \(\sqrt{2} + \sqrt{2} = \sqrt{2}\) which is not true. Therefore, option (a) is incorrect.

Step 2 Evaluating option (b)

Here, y = 0. Substituting this value into the equation, we get \(\sin x + \cos x = \sqrt{0 + \frac{1}{0}}\) which is undefined as we are trying to divide by zero. Therefore, option (b) is incorrect.

Step 3 Evaluating option (c)

Here, y = 2. We don't have a particular value for x in this case. However since we know from trigonometry that the maximum value of \(\sin x + \cos x\) is \(\sqrt{2}\), and \(2 > \sqrt{2}\), this makes the equation invalid for any value of x. Therefore, option (c) is incorrect.

Step 4 Evaluating option (d)

Here, x = \(\frac{3\pi}{4}\). Substituting this value into the equation, we get \(\sin(\frac{3\pi}{4}) + \cos(\frac{3\pi}{4}) = \sqrt{y+\frac{1}{y}}\). Solving this, which leads to a complicated expression and does not give a specific value for y without having multiple possibilities, or breaking the boundaries of the problem. Therefore, option (d) also is incorrect.

Key Concepts

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold true for every value of the involved variables. These identities help in simplifying complex trigonometric expressions or solving trigonometric equations. Some important trigonometric identities include:

• Pythagorean identities, such as \( \sin^2 x + \cos^2 x = 1 \).
• Angle sum and difference identities, like \( \sin(a \pm b) = \sin a \cos b \pm \cos a \sin b \).
• Double angle formulas, e.g., \( \sin 2x = 2 \sin x \cos x \).

In solving this exercise, we see the use of specific values as substitutions in identities to evaluate the equation. This is done step by step by plugging respective angles and comparing the trigonometric results with the identity \( \sin x + \cos x \). Recognizing these identities allows us to simplify and solve equations more efficiently.

Interval Analysis

Interval analysis involves determining the range of solutions or evaluating expressions within a specific boundary of values. Here, the interval \([0, \pi] \) is crucial because it restricts the values \( x \) can take.The benefit of interval analysis is that it narrows down our search when solving equations. By applying it:

• It limits the possible values for \( x \) to those meaningful for the trigonometric functions.
• We understand that the maximum value of \( \sin x + \cos x \), which is \( \sqrt{2} \), only occurs at specific points within the interval.

Exploring this interval allows us to discard irrelevant or impossible solutions quickly, as seen when evaluating each given option against this interval to eliminate incorrect choices. Interval checks become vital when using trigonometric identities within restricted ranges.

Function Evaluation

Function evaluation in the context of this problem involves calculating the values of trigonometric functions at specific angles and comparing them to given expressions. This step-by-step approach helps ensure we determine the validity of each potential solution correctly.When evaluating functions here:

• For each given option, substitute \( x \) and \( y \) values into the expression \( \sin x + \cos x = \sqrt{y + \frac{1}{y}} \).
• The use of known trigonometric values like \( \sin \frac{\pi}{4} \) and \( \cos \frac{\pi}{4} \) becomes essential.
• By comparing calculated results against set options, incorrect assumptions or undefined results (such as dividing by zero) are identified and ruled out.

Correctly evaluating functions for these specific cases not only provides the correct answer but also develops a deeper understanding of how trigonometric functions behave on given intervals.