Let fk(x) = 1/k(sinkx + coskx) for k = 1, 2, 3, .... Then for all x ∈ R, the value of f4(x) – f6(x)

Q: Wait, why do we reject the negative root here? Or why can't we just use energy conservation?

That is a great question, but remember that this is a trigonometric identity problem, not a mechanics problem! We are not solving for $x$ here; we are simplifying an expression. There are no roots to reject because we are not dealing with a quadratic equation in $x$. We are using algebraic identities to simplify the expression into a constant value.

Q: How do I know which identity to use for $\sin^n x + \cos^n x$?

The key is to always relate the expression back to the fundamental identity $\sin^2 x + \cos^2 x = 1$. For even powers like $4$ and $6$, you can treat them as powers of $(\sin^2 x)$ and $(\cos^2 x)$. For $n=4$, use the square identity. For $n=6$, use the cube identity. It is all about breaking down the higher powers into the fundamental building block of $\sin^2 x + \cos^2 x$.

Q: Is there a general formula for $f_k(x)$?

Yes, there is! While we only solved for $k=4$ and $k=6$, you can generalize this for any even $k$. The expression $\sin^k x + \cos^k x$ can always be expanded using the binomial theorem or Newton's sums, but for JEE, focusing on $k=2, 4, 6$ is usually sufficient to identify the pattern of how the terms cancel out.

Question

Let $f_k(x) = \frac{1}{k}(\sin^k x + \cos^k x)$ for $k = 1, 2, 3, \dots$. Then for all $x \in R$, the value of $f_4(x) - f_6(x)$ is equal to :

Options:

(A) $\frac{5}{12}$

(B) $\frac{-1}{12}$

(C) $\frac{1}{4}$

(D) $\frac{1}{12}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow explorers of the mathematical universe. Today, we are tackling a problem that might look like a daunting trigonometric beast, but beneath its surface lies a beautiful, elegant simplicity.

We are given the function $f_k(x) = \frac{1}{k}(\sin^k x + \cos^k x)$ and asked to evaluate $f_4(x) - f_6(x)$.

Before we even touch our pens to paper, look at the options. They are all constants. This is the first, most vital lesson in competitive exams: if the options are constants, the variable $x$ is a ghost. It is destined to vanish. Our mission is to orchestrate its disappearance.

### Taming the Fourth Power

Let us start with $f_4(x)$. By substituting $k=4$, we get:

$$f_4(x) = \frac{1}{4}(\sin^4 x + \cos^4 x)$$

The fourth power looks intimidating, but we have a secret weapon: algebraic identities. We know that $a^2 + b^2 = (a+b)^2 - 2ab$.

If we let $a = \sin^2 x$ and $b = \cos^2 x$, then $a^2 + b^2 = \sin^4 x + \cos^4 x$. Substituting this into our identity, we get:

$$\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x$$

Since $\sin^2 x + \cos^2 x = 1$, this simplifies to $1^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x$. Thus:

$$f_4(x) = \frac{1}{4}(1 - 2\sin^2 x \cos^2 x) = \frac{1}{4} - \frac{1}{2}\sin^2 x \cos^2 x$$

### Conquering the Sixth Power

Now, we turn our attention to $f_6(x)$. Substituting $k=6$, we have:

$$f_6(x) = \frac{1}{6}(\sin^6 x + \cos^6 x)$$

Here, we use the sum of cubes identity: $a^3 + b^3 = (a+b)^3 - 3ab(a+b)$. Again, let $a = \sin^2 x$ and $b = \cos^2 x$.

Then $a^3 + b^3 = \sin^6 x + \cos^6 x$. Applying the identity, we get:

$$\sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)^3 - 3\sin^2 x \cos^2 x(\sin^2 x + \cos^2 x)$$

With $\sin^2 x + \cos^2 x = 1$, this collapses to $1^3 - 3\sin^2 x \cos^2 x(1) = 1 - 3\sin^2 x \cos^2 x$. Therefore:

$$f_6(x) = \frac{1}{6}(1 - 3\sin^2 x \cos^2 x) = \frac{1}{6} - \frac{1}{2}\sin^2 x \cos^2 x$$

### The Grand Cancellation

Now, the moment of truth. We need to evaluate $f_4(x) - f_6(x)$. Substituting our derived expressions, we get:

$$f_4(x) - f_6(x) = \left( \frac{1}{4} - \frac{1}{2}\sin^2 x \cos^2 x \right) - \left( \frac{1}{6} - \frac{1}{2}\sin^2 x \cos^2 x \right)$$

Watch closely as we distribute the negative sign:

$$\frac{1}{4} - \frac{1}{2}\sin^2 x \cos^2 x - \frac{1}{6} + \frac{1}{2}\sin^2 x \cos^2 x$$

The terms $-\frac{1}{2}\sin^2 x \cos^2 x$ and $+\frac{1}{2}\sin^2 x \cos^2 x$ are perfect opposites. They cancel out completely, leaving us with:

$$\frac{1}{4} - \frac{1}{6} = \frac{3}{12} - \frac{2}{12} = \frac{1}{12}$$

The complexity has dissolved, leaving behind a simple, elegant fraction. The final answer is **$\frac{1}{12}$**.

Animated Solution for Mathematics - Trigonometry: Let $f_{k} (x) = \frac{1}{k} (sin^{k} x + cos^{k} x)$ for $k = 1, 2, 3, \dots$ . Then for all $x \in R$ , the value of $f_{4} (x) - f_{6} (x)$ is equal to :

Select Answer:

Visualized Solution

Understanding the Function $f_{k} (x)$

Defining $f_{4} (x)$

Algebraic Identity for $a^{4} + b^{4}$

Applying Identity to $sin^{4} x + cos^{4} x$

Final Expression for $f_{4} (x)$

Defining $f_{6} (x)$

Algebraic Identity for $a^{3} + b^{3}$

Applying Identity to $sin^{6} x + cos^{6} x$

Final Expression for $f_{6} (x)$

Setting up the Subtraction

The Magic of Cancellation

Final Arithmetic Calculation

Conclusion and Key Takeaway

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

Taming the Fourth Power

Conquering the Sixth Power

The Grand Cancellation

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Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let fk​(x)=k1​(sinkx+coskx) for k=1,2,3,…. Then for all x∈R, the value of f4​(x)−f6​(x) is equal to :

Select Answer:

Visualized Solution

Understanding the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } fk​(x)

Defining .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f4​(x)

Algebraic Identity for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a4+b4

Applying Identity to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin4x+cos4x

Final Expression for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f4​(x)

Defining .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f6​(x)

Algebraic Identity for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a3+b3

Applying Identity to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin6x+cos6x

Final Expression for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f6​(x)