Problem 10 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Solve the following trigonometric equations: \(\sin ^{4} x+\cos ^{4} x=\frac{7}{2} \sin x \cos x\)

Step-by-Step Solution

Answer

The equation simplifies to \(2\sin^2 x\cos^2x + \frac{7}{2} \sin x \cos x - 1 = 0\), the actual values of x will depend upon the domain.

Step 1: Identify Pythagorean Identity

Observe the given equation \(\sin ^{4} x+\cos ^{4} x=\frac{7}{2} \sin x \cos x\) and notice that \(sin^4x\) and \(cos^4x\) can be rewritten in terms of \((sin^2x)^2\) and \((cos^2x)^2\). This leads to the use of the well-known Pythagorean identity: \(\sin^2 x + \cos^2 x = 1\).

Step 2: Substitute Pythagorean Identity

We substitute \((\sin^2 x + \cos^2 x)^2 - 2\sin^2 x\cos^2x\) in place of \(\sin^4 x + \cos^4 x\) using the formula \((a+b)^2 = a^2 + 2ab + b^2\) and we have \((\sin^2 x + \cos^2 x)^2 - 2\sin^2 x\cos^2x = \frac{7}{2} \sin x \cos x\). As \(\sin^2 x + \cos^2 x = 1\), the equation simplifies to \(1 - 2\sin^2 x\cos^2x = \frac{7}{2} \sin x \cos x\)

Step 3: Solve for x

Next, we solve for the variable x. Put the equation into standard form: \(2\sin^2 x\cos^2x + \frac{7}{2} \sin x \cos x - 1 = 0\) . We can then factor and solve this quadratic equation.

Key Concepts

Pythagorean identity

The Pythagorean identity is a fundamental concept in trigonometry. It states that for any angle \(x\), the square of the sine of \(x\) added to the square of the cosine of \(x\) equals one: \(\sin^2 x + \cos^2 x = 1\). This identity is derived from the Pythagorean theorem in relation to the unit circle, where the radius is one.

• Imagine a right-angled triangle inscribed in a circle with radius 1.
• The height of the triangle is \(\sin(x)\) and the base is \(\cos(x)\).
• According to the Pythagorean theorem, the sum of the squares of these two legs is equal to the square of the radius, which is 1.

This identity is crucial because it allows us to transform and simplify many trigonometric equations. In our problem, by utilizing this identity to substitute certain expressions, we simplify the equation significantly, making it much more manageable.

quadratic equations

Quadratic equations often appear in trigonometry under disguised forms. They are algebraic equations where the highest power of the unknown variable is 2. In standard form, a quadratic equation looks like: \(ax^2 + bx + c = 0\). To solve these equations, there are several methods:

• Factoring, where the equation is expressed as the product of two binomials.
• The quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), a direct way to find the solutions.
• Completing the square, a method where a perfect square trinomial is formed on one side of the equation.

For the problem at hand, rewriting the equation led to a form that resembles a quadratic equation, allowing us to apply these solving techniques. This approach transforms a trigonometric problem into an algebraic one, broadening the toolkit we can use to find solutions.

trigonometric identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the involved variables. They are essential tools for simplifying and solving trigonometric equations. Beyond the Pythagorean identity, there are several other identities that help manipulate expressions:

• Angle Sum and Difference Identities for calculating the sine or cosine of an angle that is a sum or difference of two known angles.
• Double Angle Identities provide expressions for trigonometric functions of angles like \(2x\), such as \(\sin(2x) = 2\sin(x)\cos(x)\).
• Half Angle Identities help in finding the trigonometric values for half of a given angle.

The power of these identities lies in their ability to convert complex trigonometric expressions into algebraically simpler ones, enabling us to solve otherwise challenging problems. In solving the original equation, these identities allow the simplification we used to substitute and align the equation with known algebraic solutions like quadratics.