Problem 12 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The solution of \(\sin ^{8} x+\cos ^{8} x=\frac{17}{32}\) is (A) \(\frac{n \pi}{2} \pm \frac{\pi}{8}\) (B) \(n \pi \pm \frac{\pi}{4}\) (C) \(n \pi \pm \frac{\pi}{0}\) (D) no solution

Step-by-Step Solution

Answer

The given equation has no real solution (option D).

Step 1: Simplify Identity

The given identity is \( \sin^8 x + \cos^8 x = \frac{17}{32} \). We will simplify it using the identity \( a^8 + b^8 \) where \( a = \sin x \) and \( b = \cos x \). This can be expressed as \( (a^2 + b^2)^4 - 4a^2b^2(a^2 + b^2)^2 + 6a^4b^4 \). Since \( a^2 + b^2 = 1 \), the equation reduces to \( 1 - 4a^2b^2 + 6a^4b^4 \).

Step 2: Substitute Identity into Equation

Substitute \( 4a^2b^2 = 4\sin^2 x \cos^2 x = \sin^2 2x \) into the equation, giving us \( 1 - \sin^2 2x + 6(\sin^2 x \cos^2 x)^2 = \frac{17}{32} \).

Step 3: Solve for Simplified Identity

Let \( t = \sin^2 2x \), then the reduced equation is \( 1 - t + \frac{3t^2}{2} = \frac{17}{32} \). Rearrange and solve for \( t \): \( \frac{3t^2}{2} - t + \frac{15}{32} = 0 \).

Step 4: Solve the Quadratic Equation

The quadratic equation is \( 3t^2 - 16t + 15 = 0 \). Solving this using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3, b = -16, c = 15 \), gives \( t = \frac{16 \pm \sqrt{256 - 180}}{6} \), or \( t = \frac{16 \pm \sqrt{76}}{6} \). Calculate \( \sqrt{76} = 2\sqrt{19} \), yielding the roots \( t = \frac{16 \pm 2\sqrt{19}}{6} \).

Step 5: Evaluate Possible Values

The possible roots \( t = \sin^2 2x \) should lie between 0 and 1. Calculating approximately, the first root \( \frac{16 + 2\sqrt{19}}{6} \approx 6.33 \) and the second \( \frac{16 - 2\sqrt{19}}{6} \approx 1.67 \), neither of which are in the interval between 0 and 1.

Key Concepts

Quadratic Formula

The quadratic formula is a fundamental tool in algebra and is used to solve quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). In this form, \( a \), \( b \), and \( c \) are constants, and \( x \) represents the variable we are solving for.
The quadratic formula is given by:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This formula allows us to find the roots of the quadratic equation, where the roots are the values of \( x \) that satisfy the equation.
In the solution provided above, the quadratic equation \( 3t^2 - 16t + 15 = 0 \) is addressed using the quadratic formula. Here, \( a = 3 \), \( b = -16 \), and \( c = 15 \).
Applying the formula:

• The discriminant is calculated as \( b^2 - 4ac = 256 - 180 = 76 \).
• The roots then become \( t = \frac{16 \pm \sqrt{76}}{6} \).

Understanding how to use the quadratic formula is crucial for solving problems where direct factoring isn't obvious or possible.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value within their domain. These identities are essential for simplifying trigonometric expressions and solving trigonometric equations.
A key identity used in the solution is the Pythagorean identity, \( \sin^2 x + \cos^2 x = 1 \), which simplifies expressions involving higher powers of sine and cosine. In this problem, the identity helps to reframe \( \sin^8 x + \cos^8 x \) by recognizing \( a = \sin x \) and \( b = \cos x \) and writing \( (a^2 + b^2)^4 = 1^4 = 1 \).
Another trick from trigonometry is transforming products of sine and cosine, like \( 4\sin^2 x \cos^2 x \), into single terms using the double angle identity, \( \sin 2x = 2\sin x\cos x \). This identity significantly simplifies the problem and reduces it to a simpler one that makes it manageable with algebraic methods.
Mastering these identities allows students to manipulate and simplify complex trigonometric expressions easily.

Trigonometric Functions

Trigonometric functions relate the angles of a triangle to the lengths of its sides and are fundamental in the study of periodic phenomena. The primary functions are sine (\( \sin \)), cosine (\( \cos \)), and tangent (\( \tan \)), with sine and cosine being the focus here.
These functions are periodic, meaning they repeat their values in regular intervals. For sine and cosine, this interval is \( 2\pi \), directly related to the circumference of a unit circle.
In problems like the one above, understanding the properties of these functions is crucial:

• \( \sin^2 x \) and \( \cos^2 x \) are bounded between 0 and 1, conveying that trigonometric outcomes are always normalized in terms of maximum and minimum values.
• When squared, the maximum value for \( \sin x \) and \( \cos x \) remains constrained within 0 and 1.

In our problem, evaluating \( \sin^2 2x \) required knowing that these values couldn't exceed 1. Whether solving equations or modeling waveforms, understanding these limits is essential for accurate interpretation and solution of trigonometric problems.