$\sin^{4}x+\cos^{4}x$ I should rewrite this expression into a new form to plot the function.
\begin{align} & = (\sin^2x)(\sin^2x) - (\cos^2x)(\cos^2x) \\ & = (\sin^2x)^2 - (\cos^2x)^2 \\ & = (\sin^2x - \cos^2x)(\sin^2x + \cos^2x) \\ & = (\sin^2x - \cos^2x)(1) \longrightarrow\, = \sin^2x - \cos^2x \end{align}
Is that true?
\begin{align} \sin^4 x +\cos^4 x&=\sin^4 x +2\sin^2x\cos^2 x+\cos^4 x - 2\sin^2x\cos^2 x\\ &=(\sin^2x+\cos^2 x)^2-2\sin^2x\cos^2 x\\ &=1^2-\frac{1}{2}(2\sin x\cos x)^2\\ &=1-\frac{1}{2}\sin^2 (2x)\\ &=1-\frac{1}{2}\left(\frac{1-\cos 4x}{2}\right)\\ &=\frac{3}{4}+\frac{1}{4}\cos 4x \end{align}
Let $$\displaystyle y=\sin^4 x+\cos^4 x = \left(\sin^2 x+\cos^2 x\right)^2-2\sin^2 x\cdot \cos^2 x = 1-\frac{1}{2}\left(2\sin x\cdot \cos x\right)^2$$
Now using $$ \sin 2A = 2\sin A\cos A$$
So, we get $$\displaystyle y=1-\frac{1}{2}\sin^2 2x$$
Expand in terms of complex exponentials.
$$\sin^4 x + \cos^4 x = \left( \frac{e^{ix} - e^{-ix}}{2i} \right)^4 + \left( \frac{e^{ix} + e^{-ix}}{2} \right)^4$$
Notice that $i^4 = +1$, so we get
$$\sin^4 x + \cos^4 x = \frac{1}{16} \left( 2e^{4ix} + 2 e^{-4ix} + 12 \right)$$
where we use the relation $(a+b)^4 = a^4 + 4 a^3 b + 6 a^2 b^2 + 4 ab^3 + b^4$. The terms of the form $a^3 b$ and $ab^3$ all cancel by addition.
This leaves us with a final result:
$$\sin^4 x + \cos^4 x = \frac{4}{16} \left(\frac{e^{4ix} + e^{-4ix}}{2} \right) + \frac{12}{16} = \frac{3}{4} + \frac{1}{4} \cos 4x$$
Note that $a^2 + b^2 = (a+b)^2 - 2ab$
$$(\sin^2 x)^2 + (\cos^2 x)^2 = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x\cos^2 x =(\sin^2 x + \cos^2 x)^2 - 2(\sin x\cos x)^2 = \\ 1 -\frac{ \sin^2 2x}{2}$$
Note the following results:
$$ \sin^2 x + \cos^2 x = 1$$
$$ \sin x \cos x = \frac{\sin 2x}{2}$$
If you want to express in functions of higher frequencies like this $$\sum_{k=0}^N \sin(kx) + \cos(kx)$$ Then you can use the Fourier transform together with convolution theorem. This will work out for any sum of powers of cos and sin, even $\sin^{666}(x)$.
