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

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.

• Thanks to all of you for your effort and I know that your answers are right but I wanna know, in which step I have done something wrong. Why is my answer coming different? – Avi Feb 25 '17 at 3:47
• In line (9), if that "$-$" had been a "$+$", your answer would match the expected one. As it turns out, the sign error was introduced in line (5). From line (4), you should have, for instance, ... $$3 - 2 \sin^2 x\;\to\;3-2(1-\cos^2x) \;\to\; 3 - 2 \color{red}{+} 2 \cos^2 x \;\to\; 1 \color{red}{+} 2 \cos^2 x$$ Likewise, line (5) should have $1\color{red}{+}2\sin^2 x$. These sign changes carry through to line (9). – Blue Feb 25 '17 at 18:32

HINT:

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

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

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 :(

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