If $f_k(x)=\frac{1}{k}\left (\sin^kx +\cos^kx\right)$, then $f_4(x)-f_6(x)=\;?$

Question

I arrived to this question while solving a question paper. The question is as follows:

If $f_k(x)=\frac{1}{k}\left(\sin^kx + \cos^kx\right)$, where $x$ belongs to $\mathbb{R}$ and $k>1$, then $f_4(x)-f_6(x)=?$

I started as

$$\begin{align} f_4(x)-f_6(x)&=\frac{1}{4}(\sin^4x + \cos^4x) - \frac{1}{6}(\sin^6x + \cos^6x) \tag{1}\\[4pt] &=\frac{3}{12}\sin^4x + \frac{3}{12}\cos^4x - \frac{2}{12}\sin^6x - \frac{2}{12}\cos^6x \tag{2}\\[4pt] &=\frac{1}{12}\left(3\sin^4x + 3\cos^4x - 2\sin^6x - 2\cos^6x\right) \tag{3}\\[4pt] &=\frac{1}{12}\left[\sin^4x\left(3-2\sin^2x\right) + \cos^4x\left(3-2\cos^2x\right)\right] \tag{4}\\[4pt] &=\frac{1}{12}\left[\sin^4x\left(1-2\cos^2x\right) + \cos^4x\left(1-2\sin^2x\right)\right] \tag{5} \\[4pt] &\qquad\quad \text{(substituting $\sin^2x=1-\cos^2x$ and $\cos^2x=1-\sin^2x$)} \\[4pt] &=\frac{1}{12}\left(\sin^4x-2\cos^2x\sin^4x+\cos^4x-2\sin^2x\cos^4x\right) \tag{6} \\[4pt] &=\frac{1}{12}\left[\sin^4x+\cos^4x-2\cos^2x\sin^2x\left(\sin^2x+\cos^2x\right)\right] \tag{7} \\[4pt] &=\frac{1}{12}\left(\sin^4x+\cos^4x-2\cos^2x\sin^2x\right) \tag{8} \\[4pt] &\qquad\quad\text{(because $\sin^2x+\cos^2x=1$)} \\[4pt] &=\frac{1}{12}\left(\cos^2x-\sin^2x\right)^2 \tag{9} \\[4pt] &=\frac{1}{12}\cos^2(2x) \tag{10}\\[4pt] &\qquad\quad\text{(because $\cos^2x-\sin^2x=\cos2x$)} \end{align}$$

Hence the answer should be ...

$$f_4(x)-f_6(x)=\frac{1}{12}\cos^2(2x)$$

... but the answer given was $\frac{1}{12}$.

I know this might be a very simple question but trying many a times also didn't gave me the right answer. Please tell me where I am doing wrong.

Answer 1

Hint. Let $x=\frac{\pi}{4}$

In your answer $f4(x)-f(6x) = 0$

By brute-forse, $f4(x)-f6(x) = \frac{0.5}{4} - \frac{0.25}{6} = \frac{1}{8}-\frac{1}{24} =\frac{1}{12}$, so, you make a typo :(

Answer 2

Hint: let $f(x):=f_4(x)-f_6(x)$. Then show that $f'(x)=0$ for all $x$. Hence $f$ is constant. Furthermore: $f(0)=\frac{1}{12}$

Answer 3

HINT:

$$\sin^6x+\cos^6x=(\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x(\sin^2x+\cos^2x)=1-3\sin^2x\cos^2x$$

$$\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-2\sin^2x\cos^2x$$

Answer 4

‘Tis I have a better solution to solve this math problem...

Let $a = \cos^2 x$ and $b = \sin^2 x,$ so $a + b = 1.$ Then [(a + b)^2 = a^2 + 2ab + b^2 = 1,]so $a^2 + b^2 = 1 - 2ab.$ Also, [(a + b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3 = 1,]so \begin{align*} a^3 + b^3 &= 1 - (3a^2 b + 3ab^2) \\ &= 1 - 3ab(a + b) \\ &= 1 - 3ab. \end{align*}Therefore, \begin{align*} f_4(x) - f_6(x) &= \frac{\sin^4 x + \cos^4 x}{4} - \frac{\sin^6 x + \cos^6 x}{6} \\ &= \frac{a^2 + b^2}{4} - \frac{a^3 + b^3}{6} \\ &= \frac{1 - 2ab}{4} - \frac{1 - 3ab}{6} \\ &= \boxed{\frac{1}{12}}. \end{align*}

BTW, I solved this