Problem 44 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\sqrt{\cos 2 x}+\sqrt{1+\sin 2 x}=\sqrt{\sin x+\cos x}\) if (A) \(x=2 n \pi\) (B) \(x=n \pi-\frac{\pi}{4}\) (C) \(\sin x+\cos x=0\) (D) \(x=n \pi\)

Step-by-Step Solution

Answer

The correct option is (C): \(\sin x + \cos x = 0\).

Step 1: Simplify the Equation

First, simplify the given equation: \[ \sqrt{\cos 2x} + \sqrt{1 + \sin 2x} = \sqrt{\sin x + \cos x} \]Recall the double angle identities: \(\cos 2x = \cos^2 x - \sin^2 x\) and \(\sin 2x = 2 \sin x \cos x\). Use these identities to rewrite the left side.

Step 2: Analyze the Expression

Express \(\cos 2x\) and \(1 + \sin 2x\):\[ \cos 2x = 1 - 2 \sin^2 x \]\[ 1 + \sin 2x = 1 + 2 \sin x \cos x = (\sin x + \cos x)^2 \]Substitute these into the original equation to analyze further.

Step 3: Evaluate Option A

If \(x = 2n\pi\), then \(\cos x = \cos 2n\pi = 1\) and \(\sin x = \sin 2n\pi = 0\). Thus, \(\sin x + \cos x = 1\).Substituting into the original expression gives: \[ \sqrt{\cos 4n\pi} + \sqrt{1 + \sin 4n\pi} = \sqrt{1}\]Since \(\cos 4n\pi = 1\) and \(\sin 4n\pi = 0\), this transforms to:\[ 1 + 1 = 1 \]This does not hold true.

Step 4: Evaluate Option B

If \(x = n\pi - \frac{\pi}{4}\), then \(\sin x + \cos x = \sin(n\pi - \frac{\pi}{4}) + \cos(n\pi - \frac{\pi}{4})\).Use the identities:\[ \sin(n\pi - \frac{\pi}{4}) = (-1)^n \frac{\sqrt{2}}{2} \]\[ \cos(n\pi - \frac{\pi}{4}) = (-1)^{n+1} \frac{\sqrt{2}}{2} \]Thus, \(\sin x + \cos x = 0\), corresponding to option C.

Step 5: Verify Option C

If \(\sin x + \cos x = 0\), then \(x = n\pi - \frac{\pi}{4}\), matching result from step 4.The equation holds as both sides simplify to zero, confirming a valid solution pathway.

Step 6: Evaluate Option D

If \(x = n\pi\), then both \(\sin x = 0\) and \(\cos x = (-1)^n\). So, \(\sin x + \cos x = (-1)^n \).Substitute into the equation:\[\sqrt{\cos 2n\pi} + \sqrt{1 + 0} = \sqrt{(-1)^n} \]This becomes:\[ 1 + 1 = \sqrt{(-1)^n} \]This does not always hold true, as the right side cannot consistently equal 2.

Key Concepts

Double Angle Identities

When dealing with trigonometric identities, the double angle identities are key tools in simplifying and solving equations. Double angle identities express trigonometric functions of double angles like \( 2x \) in terms of functions of \( x \). They are particularly helpful for re-expressing the terms in a trigonometric equation to make it easier to solve. Here are two of the most commonly used double angle identities:

• \( \cos 2x = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x = 2 \cos^2 x - 1 \)
• \( \sin 2x = 2 \sin x \cos x \)

These identities allow us to break down the initial factors into forms that are more manageable and sometimes more recognizable for further manipulation.
In the context of our exercise, these identities help convert the terms \( \sqrt{\cos 2x} \) and \( \sqrt{1 + \sin 2x} \) into manageable expressions that facilitate exploration of valid solutions for \( x \). The process involves algebraic manipulation to simplify the equations and identify opportunities to equate both sides conveniently.

Trigonometric Equations

Trigonometric equations contain trigonometric functions and are solved to find the values of the variable, often angles, that satisfy the equation. Solving these equations can involve several methods, including the use of identities, algebraic manipulation, and considering periodic properties.
In our exercise, solving the equation \( \sqrt{\cos 2x} + \sqrt{1 + \sin 2x} = \sqrt{\sin x + \cos x} \) involves evaluating the equation at certain values of \( x \), like \( 2n\pi \), \( n\pi - \frac{\pi}{4} \), and \( n\pi \). Each option involves substituting specific values of \( x \) and checking if both sides of the equation equalize.

• For \( x = 2n\pi \), calculations showed that the left side did not equal the right side.
• For \( x = n\pi - \frac{\pi}{4} \), both sides simplified to zero, suggesting a valid equation.
• For \( x = n\pi \), the solution did not consistently hold, as the form changed depending on \( n \).

This process emphasizes the need to understand the properties of trigonometric functions and apply them correctly to solve the equation effectively by matching the trigonometric expressions on both sides.

Trigonometric Simplification

Trigonometric simplification plays a crucial role in making complex expressions more manageable, ultimately aiding in solving equations or verifying identities. Simplification involves rewriting trigonometric expressions using known identities and algebraic transformations.
In our exercise, the initial complex equation \( \sqrt{\cos 2x} + \sqrt{1 + \sin 2x} = \sqrt{\sin x + \cos x} \) is simplified by employing double angle identities and expressing certain terms in simpler forms:

• \( \cos 2x \) is expressed as \( 1 - 2 \sin^2 x \).
• \( 1 + \sin 2x \) is simplified to \( (\sin x + \cos x)^2 \).

These transformations make it easier to analyze the equation and evaluate it under different scenarios, checking which conditions hold true for specific values of \( x \).
Making the expressions on each side of the equation more comparable using appropriate identities is a key step in solving trigonometric equations efficiently. It reduces the complexity and allows us to observe patterns or symmetries that suggest simplifications or solutions. This principle showcases how fundamental trigonometric identities form the backbone of solving and simplifying trigonometric equations.