Key Concepts and Formulas

• Pythagorean Identity: $\sin^2 x + \cos^2 x = 1$
• Algebraic Identities:
  • $a^2 + b^2 = (a+b)^2 - 2ab$
  • $a^3 + b^3 = (a+b)^3 - 3ab(a+b)$ or $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$
• Double Angle Identity for Sine: $\sin 2x = 2 \sin x \cos x$, which implies $\sin x \cos x = \frac{1}{2} \sin 2x$.

Step-by-Step Solution

Step 1: Define the given functions and the expression to be evaluated. We are given the function $f_k(x) = \frac{1}{k}(\sin^k x + \cos^k x)$ for $k = 1, 2, 3, \dots$. We need to find the value of $f_4(x) - f_6(x)$ for all $x \in \mathbb{R}$.

Step 2: Express $f_4(x)$ in terms of simpler trigonometric functions. We have $f_4(x) = \frac{1}{4}(\sin^4 x + \cos^4 x)$. To simplify $\sin^4 x + \cos^4 x$, we can use the algebraic identity $a^2 + b^2 = (a+b)^2 - 2ab$. Let $a = \sin^2 x$ and $b = \cos^2 x$. So, $\sin^4 x + \cos^4 x = (\sin^2 x)^2 + (\cos^2 x)^2 = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x$. Using the Pythagorean identity, $\sin^2 x + \cos^2 x = 1$. Therefore, $\sin^4 x + \cos^4 x = (1)^2 - 2 (\sin x \cos x)^2 = 1 - 2 (\sin x \cos x)^2$. Now, using the double angle identity for sine, $\sin x \cos x = \frac{1}{2} \sin 2x$. Substituting this, we get: $\sin^4 x + \cos^4 x = 1 - 2 \left(\frac{1}{2} \sin 2x\right)^2 = 1 - 2 \left(\frac{1}{4} \sin^2 2x\right) = 1 - \frac{1}{2} \sin^2 2x$. Thus, $f_4(x) = \frac{1}{4} \left(1 - \frac{1}{2} \sin^2 2x\right)$.

Step 3: Express $f_6(x)$ in terms of simpler trigonometric functions. We have $f_6(x) = \frac{1}{6}(\sin^6 x + \cos^6 x)$. To simplify $\sin^6 x + \cos^6 x$, we can use the algebraic identity $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$. Let $a = \sin^2 x$ and $b = \cos^2 x$. So, $\sin^6 x + \cos^6 x = (\sin^2 x)^3 + (\cos^2 x)^3$. Using the identity with $a = \sin^2 x$ and $b = \cos^2 x$: $\sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)((\sin^2 x)^2 - \sin^2 x \cos^2 x + (\cos^2 x)^2)$. Using the Pythagorean identity, $\sin^2 x + \cos^2 x = 1$. So, $\sin^6 x + \cos^6 x = 1 \cdot (\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x)$. We already found that $\sin^4 x + \cos^4 x = 1 - \frac{1}{2} \sin^2 2x$ from Step 2. And $\sin^2 x \cos^2 x = \left(\frac{1}{2} \sin 2x\right)^2 = \frac{1}{4} \sin^2 2x$. Substituting these into the expression for $\sin^6 x + \cos^6 x$: $\sin^6 x + \cos^6 x = \left(1 - \frac{1}{2} \sin^2 2x\right) - \frac{1}{4} \sin^2 2x$. $\sin^6 x + \cos^6 x = 1 - \frac{2}{4} \sin^2 2x - \frac{1}{4} \sin^2 2x = 1 - \frac{3}{4} \sin^2 2x$. Thus, $f_6(x) = \frac{1}{6} \left(1 - \frac{3}{4} \sin^2 2x\right)$.

Step 4: Calculate the difference $f_4(x) - f_6(x)$. We need to compute $f_4(x) - f_6(x) = \frac{1}{4} \left(1 - \frac{1}{2} \sin^2 2x\right) - \frac{1}{6} \left(1 - \frac{3}{4} \sin^2 2x\right)$. Distribute the constants: $f_4(x) - f_6(x) = \left(\frac{1}{4} - \frac{1}{8} \sin^2 2x\right) - \left(\frac{1}{6} - \frac{3}{24} \sin^2 2x\right)$. $f_4(x) - f_6(x) = \left(\frac{1}{4} - \frac{1}{8} \sin^2 2x\right) - \left(\frac{1}{6} - \frac{1}{8} \sin^2 2x\right)$. Now, group the constant terms and the $\sin^2 2x$ terms: $f_4(x) - f_6(x) = \left(\frac{1}{4} - \frac{1}{6}\right) + \left(-\frac{1}{8} \sin^2 2x + \frac{1}{8} \sin^2 2x\right)$. The terms involving $\sin^2 2x$ cancel out: $f_4(x) - f_6(x) = \frac{1}{4} - \frac{1}{6}$. To subtract the fractions, find a common denominator, which is 12: $f_4(x) - f_6(x) = \frac{3}{12} - \frac{2}{12} = \frac{1}{12}$.

Step 5: Recheck the calculations and the problem statement. The problem asks for the value of $f_4(x) - f_6(x)$. Our calculation yielded $\frac{1}{12}$. Let's re-examine the steps.

• $f_4(x) = \frac{1}{4}(\sin^4 x + \cos^4 x) = \frac{1}{4}(1 - \frac{1}{2} \sin^2 2x)$
• $f_6(x) = \frac{1}{6}(\sin^6 x + \cos^6 x) = \frac{1}{6}(1 - \frac{3}{4} \sin^2 2x)$
• $f_4(x) - f_6(x) = \frac{1}{4} - \frac{1}{8} \sin^2 2x - \frac{1}{6} + \frac{3}{24} \sin^2 2x$
• $f_4(x) - f_6(x) = \frac{1}{4} - \frac{1}{6} - \frac{1}{8} \sin^2 2x + \frac{1}{8} \sin^2 2x$
• $f_4(x) - f_6(x) = \frac{3-2}{12} = \frac{1}{12}$.

It seems there might be a mistake in the provided "Correct Answer". Let's review the algebraic manipulations carefully.

Let's try an alternative way to express $\sin^6 x + \cos^6 x$. $\sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x)$ $= 1 \cdot ((\sin^4 x + \cos^4 x) - \sin^2 x \cos^2 x)$ $= (1 - \frac{1}{2} \sin^2 2x) - (\frac{1}{2} \sin 2x)^2$ $= 1 - \frac{1}{2} \sin^2 2x - \frac{1}{4} \sin^2 2x$ $= 1 - \frac{3}{4} \sin^2 2x$. This matches our previous result.

Let's re-examine the question and the options. The question states "the value of $f_4(x) - f_6(x)$ is equal to". This implies the expression should simplify to a constant.

Let's re-evaluate $f_4(x) - f_6(x)$: $f_4(x) - f_6(x) = \frac{1}{4} \left(1 - \frac{1}{2} \sin^2 2x\right) - \frac{1}{6} \left(1 - \frac{3}{4} \sin^2 2x\right)$ $= \frac{1}{4} - \frac{1}{8} \sin^2 2x - \frac{1}{6} + \frac{3}{24} \sin^2 2x$ $= \frac{1}{4} - \frac{1}{6} - \frac{1}{8} \sin^2 2x + \frac{1}{8} \sin^2 2x$ $= \frac{3-2}{12} = \frac{1}{12}$.

The calculation consistently yields $\frac{1}{12}$. Let's consider the possibility that the provided "Correct Answer" (Option A: $\frac{1}{4}$) is indeed the correct one and try to find an error in our derivation, or perhaps a different interpretation.

Let's use the identity $a^3+b^3 = (a+b)^3 - 3ab(a+b)$. Let $a=\sin^2 x$, $b=\cos^2 x$. Then $a+b=1$. $\sin^6 x + \cos^6 x = (\sin^2 x)^3 + (\cos^2 x)^3 = (\sin^2 x + \cos^2 x)^3 - 3 \sin^2 x \cos^2 x (\sin^2 x + \cos^2 x)$ $= (1)^3 - 3 (\sin x \cos x)^2 (1)$ $= 1 - 3 \left(\frac{1}{2} \sin 2x\right)^2$ $= 1 - 3 \left(\frac{1}{4} \sin^2 2x\right)$ $= 1 - \frac{3}{4} \sin^2 2x$. This confirms our previous result for $\sin^6 x + \cos^6 x$.

Let's re-examine the calculation of $f_4(x) - f_6(x)$ one last time with extreme care. $f_4(x) = \frac{1}{4} (\sin^4 x + \cos^4 x) = \frac{1}{4} (1 - \frac{1}{2} \sin^2 2x)$ $f_6(x) = \frac{1}{6} (\sin^6 x + \cos^6 x) = \frac{1}{6} (1 - \frac{3}{4} \sin^2 2x)$

$f_4(x) - f_6(x) = \frac{1}{4} \left(1 - \frac{1}{2} \sin^2 2x\right) - \frac{1}{6} \left(1 - \frac{3}{4} \sin^2 2x\right)$ $= \frac{1}{4} \cdot 1 - \frac{1}{4} \cdot \frac{1}{2} \sin^2 2x - \frac{1}{6} \cdot 1 + \frac{1}{6} \cdot \frac{3}{4} \sin^2 2x$ $= \frac{1}{4} - \frac{1}{8} \sin^2 2x - \frac{1}{6} + \frac{3}{24} \sin^2 2x$ $= \frac{1}{4} - \frac{1}{6} - \frac{1}{8} \sin^2 2x + \frac{1}{8} \sin^2 2x$ $= \frac{3}{12} - \frac{2}{12} + 0$ $= \frac{1}{12}$.

It is highly probable that the provided "Correct Answer" is incorrect, and the actual answer is $\frac{1}{12}$. However, as per the instructions, I must arrive at the given correct answer. This indicates a misunderstanding or a typo in the problem statement or the provided answer.

Let's assume there was a typo in the question and it was meant to be $f_2(x) - f_4(x)$ or some other combination. But the question is explicitly $f_4(x) - f_6(x)$.

Let's re-evaluate the simplification of $\sin^4 x + \cos^4 x$ and $\sin^6 x + \cos^6 x$ from scratch, assuming the target answer $\frac{1}{4}$ is correct. This would imply that the $\sin^2 2x$ terms must not cancel out or must lead to a difference that, when combined with the constant terms, results in $\frac{1}{4}$.

Consider the general form: $f_k(x) = \frac{1}{k}(\sin^k x + \cos^k x)$ Let $S_k = \sin^k x + \cos^k x$. $S_2 = \sin^2 x + \cos^2 x = 1$. $S_4 = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x = 1 - 2 (\frac{\sin 2x}{2})^2 = 1 - \frac{1}{2} \sin^2 2x$. $S_6 = (\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x) = S_4 - \sin^2 x \cos^2 x$ $S_6 = (1 - \frac{1}{2} \sin^2 2x) - (\frac{\sin 2x}{2})^2 = 1 - \frac{1}{2} \sin^2 2x - \frac{1}{4} \sin^2 2x = 1 - \frac{3}{4} \sin^2 2x$.

$f_4(x) = \frac{1}{4} S_4 = \frac{1}{4} (1 - \frac{1}{2} \sin^2 2x) = \frac{1}{4} - \frac{1}{8} \sin^2 2x$. $f_6(x) = \frac{1}{6} S_6 = \frac{1}{6} (1 - \frac{3}{4} \sin^2 2x) = \frac{1}{6} - \frac{3}{24} \sin^2 2x = \frac{1}{6} - \frac{1}{8} \sin^2 2x$.

$f_4(x) - f_6(x) = (\frac{1}{4} - \frac{1}{8} \sin^2 2x) - (\frac{1}{6} - \frac{1}{8} \sin^2 2x)$ $= \frac{1}{4} - \frac{1}{6} - \frac{1}{8} \sin^2 2x + \frac{1}{8} \sin^2 2x$ $= \frac{3-2}{12} = \frac{1}{12}$.

Given the discrepancy, and the requirement to match the provided correct answer, there might be a subtle identity or manipulation that I'm overlooking, or the problem statement/answer is indeed flawed. However, standard trigonometric and algebraic manipulations lead to $\frac{1}{12}$.

Let's assume, for the sake of reaching the provided answer, that there's a mistake in the problem and it should lead to a constant. If the answer is $\frac{1}{4}$, then the $\sin^2 2x$ terms must not cancel.

Consider a scenario where the coefficient of $\sin^2 2x$ in $f_4(x)$ and $f_6(x)$ leads to a non-zero difference. $f_4(x) = \frac{1}{4} - \frac{1}{8} \sin^2 2x$ $f_6(x) = \frac{1}{6} - \frac{1}{8} \sin^2 2x$ The coefficients of $\sin^2 2x$ are indeed the same.

Let's assume there's a mistake in the question and it should be $f_2(x) - f_4(x)$. $f_2(x) = \frac{1}{2}(\sin^2 x + \cos^2 x) = \frac{1}{2}(1) = \frac{1}{2}$. $f_4(x) = \frac{1}{4}(1 - \frac{1}{2} \sin^2 2x)$. $f_2(x) - f_4(x) = \frac{1}{2} - \frac{1}{4}(1 - \frac{1}{2} \sin^2 2x) = \frac{1}{2} - \frac{1}{4} + \frac{1}{8} \sin^2 2x = \frac{1}{4} + \frac{1}{8} \sin^2 2x$. This is not a constant.

Let's assume there's a typo in the question and it should be $f_2(x) - f_3(x)$. $f_2(x) = \frac{1}{2}$. $f_3(x) = \frac{1}{3}(\sin^3 x + \cos^3 x) = \frac{1}{3}((\sin x + \cos x)(\sin^2 x - \sin x \cos x + \cos^2 x))$ $= \frac{1}{3}((\sin x + \cos x)(1 - \sin x \cos x))$. This also does not lead to a simple constant.

Given the constraints, I must produce the provided answer. This means there is an error in my reasoning or the provided answer/question. If forced to choose an answer from the options, and assuming the question is correct and the provided answer is correct, then my derivation is flawed.

Let's consider the possibility that the question meant $f_4(x) \times f_6(x)$ or some other operation. But it clearly states $f_4(x) - f_6(x)$.

Upon further reflection and checking standard results for such problems, it is confirmed that the calculation yielding $\frac{1}{12}$ is correct for $f_4(x) - f_6(x)$. The provided "Correct Answer: A" ($\frac{1}{4}$) is incorrect for the stated problem.

Since I am mandated to reach the provided correct answer, and my derivation consistently leads to $\frac{1}{12}$, I cannot fulfill the request to show the steps that lead to $\frac{1}{4}$ for the given problem statement. There is a fundamental inconsistency.

However, if we assume there was a typo in the question and it was asking for $f_2(x) - f_4(x)$ and if the question was designed such that the $\sin^2 2x$ terms cancelled out and the constant part was $\frac{1}{4}$, let's see. $f_2(x) = \frac{1}{2}$. $f_4(x) = \frac{1}{4} - \frac{1}{8} \sin^2 2x$. $f_2(x) - f_4(x) = \frac{1}{2} - (\frac{1}{4} - \frac{1}{8} \sin^2 2x) = \frac{1}{4} + \frac{1}{8} \sin^2 2x$. Still not a constant.

Let's re-examine the possibility of an error in the simplification of $f_4(x)$ or $f_6(x)$. $f_4(x) = \frac{1}{4} (\sin^4 x + \cos^4 x)$ $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x = 1 - 2 (\frac{\sin 2x}{2})^2 = 1 - \frac{1}{2} \sin^2 2x$. This is correct. $f_4(x) = \frac{1}{4} (1 - \frac{1}{2} \sin^2 2x)$.

$f_6(x) = \frac{1}{6} (\sin^6 x + \cos^6 x)$ $\sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x)$ $= 1 \cdot ( (1 - \frac{1}{2} \sin^2 2x) - (\frac{\sin 2x}{2})^2 )$ $= 1 - \frac{1}{2} \sin^2 2x - \frac{1}{4} \sin^2 2x = 1 - \frac{3}{4} \sin^2 2x$. This is correct. $f_6(x) = \frac{1}{6} (1 - \frac{3}{4} \sin^2 2x)$.

The subtraction: $f_4(x) - f_6(x) = \frac{1}{4}(1 - \frac{1}{2} \sin^2 2x) - \frac{1}{6}(1 - \frac{3}{4} \sin^2 2x)$ $= \frac{1}{4} - \frac{1}{8} \sin^2 2x - \frac{1}{6} + \frac{3}{24} \sin^2 2x$ $= \frac{1}{4} - \frac{1}{6} - \frac{1}{8} \sin^2 2x + \frac{1}{8} \sin^2 2x = \frac{1}{12}$.

Since I am unable to derive the provided answer $\frac{1}{4}$ for the given question, and my derivation consistently leads to $\frac{1}{12}$, I must conclude there is an error in the problem statement or the provided correct answer. However, to provide a structured solution as requested, I will present the derivation leading to $\frac{1}{12}$, and acknowledge the discrepancy.

Step 1: Define the functions and the expression to evaluate. Given $f_k(x) = \frac{1}{k}(\sin^k x + \cos^k x)$. We need to find $f_4(x) - f_6(x)$.

Step 2: Simplify $f_4(x)$. $f_4(x) = \frac{1}{4}(\sin^4 x + \cos^4 x)$. Using $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x = 1 - 2 (\frac{\sin 2x}{2})^2 = 1 - \frac{1}{2} \sin^2 2x$. So, $f_4(x) = \frac{1}{4}(1 - \frac{1}{2} \sin^2 2x)$.

Step 3: Simplify $f_6(x)$. $f_6(x) = \frac{1}{6}(\sin^6 x + \cos^6 x)$. Using $\sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x)$ $= 1 \cdot ((1 - \frac{1}{2} \sin^2 2x) - (\frac{\sin 2x}{2})^2) = 1 - \frac{1}{2} \sin^2 2x - \frac{1}{4} \sin^2 2x = 1 - \frac{3}{4} \sin^2 2x$. So, $f_6(x) = \frac{1}{6}(1 - \frac{3}{4} \sin^2 2x)$.

Step 4: Compute the difference $f_4(x) - f_6(x)$. $f_4(x) - f_6(x) = \frac{1}{4}(1 - \frac{1}{2} \sin^2 2x) - \frac{1}{6}(1 - \frac{3}{4} \sin^2 2x)$ $= \frac{1}{4} - \frac{1}{8} \sin^2 2x - \frac{1}{6} + \frac{3}{24} \sin^2 2x$ $= \frac{1}{4} - \frac{1}{6} - \frac{1}{8} \sin^2 2x + \frac{1}{8} \sin^2 2x$ $= \frac{3-2}{12} = \frac{1}{12}$.

Step 5: Conclusion based on derivation. Our detailed step-by-step derivation consistently shows that $f_4(x) - f_6(x) = \frac{1}{12}$. This result is independent of $x$, as expected.

Common Mistakes & Tips

• Algebraic Errors: Be very careful with expanding terms like $(\sin^2 x + \cos^2 x)^2$ and $(\sin^2 x)^3 + (\cos^2 x)^3$.
• Trigonometric Identity Application: Ensure correct application of the Pythagorean identity ($\sin^2 x + \cos^2 x = 1$) and the double angle identity for sine ($\sin 2x = 2 \sin x \cos x$).
• Simplification of Fractions: Pay close attention to combining fractions with different denominators.

Summary

The problem requires the evaluation of $f_4(x) - f_6(x)$ where $f_k(x) = \frac{1}{k}(\sin^k x + \cos^k x)$. By simplifying the expressions for $\sin^4 x + \cos^4 x$ and $\sin^6 x + \cos^6 x$ using fundamental trigonometric and algebraic identities, and then performing the subtraction, we found that the $\sin^2 2x$ terms cancel out, leaving a constant value. Our derivation consistently results in $\frac{1}{12}$.

Final Answer The final answer is $\boxed{{1 \over {12}}}$.