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?

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    $\begingroup$ It's going to be $K=-3/2$, not sure how to prove it besides desmos.com/calculator/glwwwu5qi1 $\endgroup$ – Jacob Claassen Oct 9 '17 at 1:23
  • $\begingroup$ You could take the derivative, set equal to zero, and solve for it. $\endgroup$ – Paul Oct 9 '17 at 1:28
  • $\begingroup$ @Paul the derivative of a function doesn't usually have much to do with the roots, especially since adding a constant (R) changes the roots without changing the derivative. $\endgroup$ – Johnathan Gross Oct 9 '17 at 1:30
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    $\begingroup$ @Paul I would like to add that I used the precalculus tag is to know if such question can be answered by using only precalculus tools, and as such avoiding derivatives. $\endgroup$ – Chris Steinbeck Bell Oct 9 '17 at 1:39
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    $\begingroup$ @JohnathanGross If the expression is independent of alpha, then we must have $dR/d\alpha =0$ for all alpha from which it is easy to solve for K. $\endgroup$ – Cyclohexanol. Oct 9 '17 at 2:16

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, $K = -3/2$. $\square$

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    $\begingroup$ Can you be more explicit without skipping the steps of expressing the equation in terms of $cos^{2}\alpha$. (In other words solving that part) I have not yet achieved the level of expertise to get the idea. $\endgroup$ – Chris Steinbeck Bell Oct 9 '17 at 1:45
  • $\begingroup$ @ChrisSteinbeckBell: How's that? $\endgroup$ – Blue Oct 9 '17 at 1:52
  • $\begingroup$ Thanks, by adding that step I got the idea. It was a bit hidden at first. $\endgroup$ – Chris Steinbeck Bell Oct 9 '17 at 1:58
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    $\begingroup$ @ChrisSteinbeckBell: Good to know. BTW: If this (or some other) answer satisfies you, be sure to tick the accompanying checkmark to "accept" it. (I notice that you have not accepted answers to most of your questions.) Accepting an answer not only grants a few more imaginary internet points to the answerer, it also removes the question from the "Unanswered" queue so that the Stack Exchange system won't recycle the question in the future. $\endgroup$ – Blue Oct 9 '17 at 2:03
  • $\begingroup$ I will take note on your recommendation. I'd like to say that I didn't had the necessary reputation to accept answers as I'm somewhat still new to this community. $\endgroup$ – Chris Steinbeck Bell Oct 9 '17 at 2:12

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 $-\dfrac{3}{2}$.

  • $\begingroup$ Can you please explain how did you come to that conclusion?. How does exactly $\alpha=\frac{\pi}{4}$ becomes into $K=-\frac{3}{2}$?. This part is not very clear. $\endgroup$ – Chris Steinbeck Bell Oct 9 '17 at 3:46
  • $\begingroup$ @ChrisSteinbeckBell I've added a bit more explanation. $\endgroup$ – Count Iblis Oct 9 '17 at 4:11
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    $\begingroup$ +1. Neat and tidy. Of course, this approach says that if such a $K$ exists, then it cannot be anything other than $-3/2$. The question statement appears to guarantee existence, so we're okay. However, if there were doubt, then one would need to verify that $K=-3/2$ eliminates the trig terms completely from the original expression. $\endgroup$ – Blue Oct 9 '17 at 4:52
  • $\begingroup$ @CountIblis Thanks for adding that part now is clear to me. Its very clever to use the fact that at certain angle both functions have the same value. 45 degrees. But does this approach works for other trigonometric functions?. $\endgroup$ – Chris Steinbeck Bell Oct 10 '17 at 3:44
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    $\begingroup$ @ChrisSteinbeckBell In general,if you just want to find the value of a parameter for which an identity holds, you can always consider special cases for the variable (in this case $\alpha$) and write down the equation that it holds for the special cases. To prove the identity requires more work. $\endgroup$ – Count Iblis Oct 11 '17 at 0:13

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 $K=\color{red}{-\frac{3}{2}}$)


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


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 $K=-\frac{3}{2}$.


$$1=(\sin^2x+\cos^2x)^3= \sin^6x+ \cos^6x+ 3\sin^2x\cos^2x(\sin^2x+\cos^2x)= \sin^6x+ \cos^6x+ 3\sin^2x\cos^2x$$ $$3\sin^2x\cos^2x=\frac{3}{2}((\sin^2x+\cos^2x)^2 - \sin^4x-\cos^4x)=\frac{3}{2}((1- \sin^4x-\cos^4x))$$ Substitute and compare with the equation you have


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