Problem 279 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \text { If } \frac{\sin ^{4} A}{a}+\frac{\cos ^{4} A}{b}=\frac{1}{a+b}, \text { prove that } \frac{\sin ^{8} A}{a^{3}}+\frac{\cos ^{8} A}{b^{3}}=\frac{1}{(a+b)^{3}} $$

Step-by-Step Solution

Answer

\(\frac{\sin^8A}{a^{3}}+\frac{\cos^8A}{b^{3}}=\frac{1}{(a+b)^{3}}\) has been proved with the given condition \(\frac{\sin^4A}{a}+\frac{\cos^4A}{b}=\frac{1}{a+b}\).

Step 1: Express Higher Powers

First, express \(sin^8A\) and \(cos^8A\) in terms of their squares. This can be done by using the formula \(x^2 = x*x\). Therefore, \(sin^8A = (sin^4A)^2\) and \(cos^8A = (cos^4A)^2\).

Step 2: Substitute Given Expressions

Next, replace \(sin^4A\) and \(cos^4A\) in the equation by their corresponding expressions from the given equation. Therefore, \(\frac{(sin^4A)^2}{a^3} = \frac{1}{a^3} * (\frac{\sin^4A}{a})^2 = \frac{1}{a^3} * \frac{1}{(a+b)^2}\) and \(\frac{(cos^4A)^2}{b^3} = \frac{1}{b^3} * (\frac{\cos^4A}{b})^2 = \frac{1}{b^3} * \frac{1}{(a+b)^2}\).

Step 3: Combine the Expressions

If you add the two expressions together, you get \(\frac{\sin^8A}{a^3}+\frac{\cos^8A}{b^3} = \frac{1}{a^3} * \frac{1}{(a+b)^2} + \frac{1}{b^3} * \frac{1}{(a+b)^2}\). This simplifies to \(\frac{1}{(a+b)^2} * (\frac{1}{a^3} + \frac{1}{b^3})\).

Step 4: Simplify the Expression

Finally, convert the fractions within the parenthesis to a common denominator, the result will be \(\frac{b+a}{a*b(a+b)^3}\). After simplification, we obtain the expression to be \(\frac{1}{(a+b)^3}\) thus proving the given equation.

Key Concepts

Sine and Cosine Functions

The sine and cosine functions are fundamental to trigonometry and are often used to describe oscillating systems. They are functions of an angle and are defined using the sides of a right triangle or the coordinates of points on a unit circle.

• The sine of an angle, denoted as \( \sin A \), is the ratio of the length of the side opposite the angle to the hypotenuse.
• The cosine of an angle, represented as \( \cos A \), is the ratio of the length of the adjacent side to the hypotenuse.

Understanding sine and cosine is crucial for solving problems involving periodic functions, waves, and rotational dynamics.
In higher mathematics, these functions can also be raised to various powers, like in this exercise, where both sine and cosine functions are raised to the fourth and eighth powers.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying expressions to make them easier to work with. In this exercise, we started by expressing \( \sin^8A \) and \( \cos^8A \) as squares of \( \sin^4A \) and \( \cos^4A \), respectively.
This step uses the basic rule: \( (x^2)^2 = x^4 \), which can be expanded to higher powers as needed. By substituting known expressions for these higher powers, you can relate a complex expression back to simpler sums or products.

• Substitute known expressions to simplify complicated terms.
• Use fractional manipulation to bring terms into a common form, making it easier to combine like terms.

This technique is invaluable for proving equations, such as showing how various trigonometric identities can be derived or converted into other equivalent forms.

Higher Powers of Trigonometric Functions

When working with higher powers of trigonometric functions, it’s essential to break them down into more manageable components. This is often achieved through the use of identities or known formulas.
For example, to handle \( \sin^8A \) and \( \cos^8A \), we expressed them in terms of squares of \( \sin^4A \) and \( \cos^4A \).

• This approach avoids the need for direct computation of root values and allows for simplification using basic algebra.
• Combining these higher powers often involves common denominators or factoring.

By expressing high powers in terms of known identities or lower powers, computations become much easier, and mathematical proofs can follow a clear and logical path.