How do you simplify an expression involving fourth and higher order trigonometric functions? The problem is as follows:
Which value of $K$ has to be in order that $R$ becomes independent from $\alpha$?.
$$R=\sin^6\alpha +\cos^6\alpha +K(\sin^4\alpha +\cos^4\alpha )$$
So far I've only come up with the idea that the solution may involve $R=0$, therefore
$$\sin^6\alpha +\cos^6\alpha +K(\sin^4\alpha +\cos^4\alpha)=0$$ 
as a result the expression becomes $0$ thus independent from $\alpha$, however the result is like this
$$-K=\frac{\sin^6\alpha +\cos^6\alpha}{\sin^4\alpha +\cos^4\alpha}$$
I am not sure if this is the right way.
Moreover, how can I simplify this expression, as it has order four and six?
 A: Let $x = \sin^2 \alpha,y = \cos^2 \alpha$. Note that $x+y = 1$ so $x^3 + y^3 = (x+y)(x^2+y^2-xy) = (x+y)((x+y)^2-3xy) = 1-3xy$
Also $x^2+y^2 = (x+y)^2-2xy = 1-2xy$
$R = (1-3xy) + K(1-2xy) = 1+K - (3+2K)xy$. So if $K = -3/2$ then $R$ will be independent of $\alpha$. If $K \neq -3/2$ then it will depend on $xy$ which then depends on $\alpha$.
Generally when you see an expression in the form of $\sin^{2n}x + \cos^{2n}x$ it helps to factor out $\sin^2 + \cos^2$ like what I did above
A: Recall that 
$$\sin^2\alpha = 1 - \cos^2\alpha$$
and express everything in terms of $\cos^2\alpha$:
$$\begin{align}
R &= \left(\;\sin^2\alpha\;\right)^3 + \left(\;\cos^2\alpha\;\right)^3+K\left(\;\left(\;\sin^2\alpha\;\right)^2+\left(\;\cos^2\alpha\;\right)^2\;\right) \\
&= \left(\;1-\cos^2\alpha\;\right)^3 + \left(\;\cos^2\alpha\;\right)^3+K\left(\;\left(\;1-\cos^2\alpha\;\right)^2+\left(\;\cos^2\alpha\;\right)^2\;\right) \\
&= \left(\;1-x\;\right)^3 + \left(\;x\;\right)^3+K\left(\;\left(\;1-x\;\right)^2+\left(\;x\;\right)^2\;\right) \qquad\text{(writing $x$ for $\cos^2\alpha$)}\\
&= 1 - 3 x + 3 x^2 + K \left(\; 1 - 2 x + 2 x^2 \;\right) \\
&= 1 +K -(3+2K) x + (3+2K) x^2
\end{align}$$
Independence from $\alpha$ translates to independence from $x$. We need a value of $K$ that causes the non-constant terms of the polynomial to vanish. Clearly, $K = -3/2$. $\square$
A: Another possible way is to start with
$$R=\sin ^6(a)+\cos ^6(a)+K \left(\sin ^4(a)+\cos ^4(a)\right)$$ Using multiple angles formalae, this rewrite as $$R=\sin ^6(a)+\cos ^6(a)+\frac{1}{4} K (\cos (4 a)+3) $$ and say that the derivative of $R$ with respect to $a$ is equal to $0$.
$$\frac{dR}{da}=6 \sin ^5(a) \cos (a)-6 \sin (a) \cos ^5(a)-K \sin (4 a)=-\frac{3}{2} \sin (4 a)-K \sin (4 a)$$ $$\frac{dR}{da}=-\frac{1}{2} (2 K+3) \sin (4 a)=0$$ Then $K=-\frac{3}{2}$.
A: For $\alpha = 0$ we have $\sin(\alpha)=0 $ and $\cos(\alpha) = 1$, therefore $R = 1+ K$ while for $\alpha = \dfrac{\pi}{4}$ we have $\cos(\alpha) = \sin(\alpha) = \dfrac{1}{\sqrt{2}}$, so $R = \dfrac{1}{4} + \dfrac{K}{2}$, therefore if there exists a value for $K$ for which the expression is independent of $\alpha$, then we must have:
$$1+ K = \frac{1}{4} + \frac{K}{2}$$
therefore $K$ must be $-\dfrac{3}{2}$.
A: $$1=(\sin^2x+\cos^2x)^3= \sin^6x+ \cos^6x+ 3\sin^2x\cos^2x(\sin^2x+\cos^2x)= \sin^6x+ \cos^6x+ 3\sin^2x\cos^2x$$
$$3\sin^2x\cos^2x=\frac{3}{2}((\sin^2x+\cos^2x)^2 - \sin^4x-\cos^4x)=\frac{3}{2}((1- \sin^4x-\cos^4x))$$
Substitute and compare with the equation you have
A: Using $ \sin^2 \alpha + \cos^2 \alpha =1$ 
\begin{eqnarray*}
\sin^4 \alpha + \cos^4 \alpha =(\sin^2 \alpha + \cos^2 \alpha)^2 -2\sin^2 \alpha  \cos^2 \alpha = 1-2\sin^2 \alpha  \cos^2 \alpha \\
\sin^6 \alpha + \cos^6 \alpha =(\sin^2 \alpha + \cos^2 \alpha)(\sin^4 \alpha  -\sin^2 \alpha \cos^2 \alpha+ \cos^4 \alpha)  = 1-3\sin^2 \alpha  \cos^2 \alpha. \\
\end{eqnarray*}
So your equation can be simplified to
\begin{eqnarray*}
R=\sin^6\alpha +\cos^6\alpha +K(\sin^4\alpha +\cos^4\alpha ) \\
=1-3\sin^2 \alpha  \cos^2 \alpha +K(1-2\sin^2 \alpha  \cos^2 \alpha) \\
=1+K-(2K+3)\sin^2 \alpha  \cos^2 \alpha \\
\end{eqnarray*}
So it is independent of $\alpha$ when $\color{blue}{2K+3=0}$. (Giving the value $K=\color{red}{-\frac{3}{2}}$)
