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The original exercise is to

Prove that

$$4(\cos^320^\circ+\sin^310^\circ)=3(\cos20^\circ+\sin10^\circ)$$

Dividing both sides by $\cos20^\circ+\sin10^\circ$ leads me to the problem in the question title.

I've tried rewriting the left side in terms of $\sin10^\circ$:

$$4\sin^410^\circ+2\sin^310^\circ-3\sin^210^\circ-\sin10^\circ+1\quad(*)$$

but there doesn't seem to be any immediate way to simplify further. I've considered replacing $x=10^\circ$ to see if there was some observation I could make about the more general polynomial $4x^4-2x^3-3x^2-x+1$ but I don't see anything particularly useful about that. Attempting to rewrite in terms of $\cos20^\circ$ seems like it would complicate things by needlessly(?) introducing square roots.

Is there a clever application of identities to arrive at the value of $\dfrac34$? I have considered

$$\cos20^\circ\sin10^\circ=\frac{\sin30^\circ-\sin10^\circ}2=\frac14-\frac12\sin10^\circ$$

which eliminates the cubic term in $(*)$, and I would have to show that

$$4\sin^410^\circ-3\sin^210^\circ+\frac12\sin10^\circ=0$$

$$4\sin^310^\circ-3\sin10^\circ+\frac12=0$$

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    $\begingroup$ I don't now how far it'll get you, but try: multiply through by 4, subtract 3 from both sides. Now you've got a polynomial of integer coefficients that you need to find the roots of. Hm... $\endgroup$ – Dan Uznanski Feb 18 at 19:15
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That last equation is trivial because $\sin 3\theta=3\sin \theta-4\sin^3\theta$.

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  • $\begingroup$ Thanks, as I made more edits I gradually came to the same conclusion :) $\endgroup$ – user170231 Feb 18 at 19:19
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A complex numbers proof of the original exercise.

By Euler's formula, we have that $$\text{Re}\left((\cos 20^\circ+i\sin 20^\circ)^3\right)=\cos(60^\circ)=\frac{1}{2}=\sin(30^\circ)=\text{Im}\left((\cos 10^\circ+i\sin 10^\circ)^3\right)$$ Hence $$\cos^3 20^\circ-3\cos 20^\circ \sin^2 20^\circ =3\cos^2 10^\circ\sin 10^\circ -\sin^3 10^\circ$$ or $$\cos^3 20^\circ-3\cos 20^\circ (1-\cos^2 20^\circ) =3(1-\sin^2 10^\circ)\sin 10^\circ -\sin^3 10^\circ $$ which implies $$4(\cos^320^\circ+\sin^310^\circ)=3(\cos20^\circ+\sin10^\circ).$$ The very same argument leads to the following generalization: if $c^\circ +s^\circ=30^\circ$ then $$4(\cos^3 c^\circ+\sin^3 s^\circ)=3(\cos c^\circ+\sin s^\circ).$$

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  • $\begingroup$ @user170231 I propose a complex numbers solution and a generalization. $\endgroup$ – Robert Z Feb 18 at 19:53
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We can use also the following way. $$\cos^220^\circ-\cos20^\circ\sin10^\circ+\sin^210^\circ=$$ $$=\frac{1}{2}(1+\cos40^{\circ})-\frac{1}{2}(\sin30^{\circ}-\sin10^{\circ})+\frac{1}{2}(1-\cos20^{\circ})=$$ $$=\frac{3}{4}+\frac{1}{2}(\cos40^{\circ}+\sin10^{\circ}-\cos20^{\circ})=$$ $$=\frac{3}{4}-\frac{1}{2}(\sin10^{\circ}-2\sin30^{\circ}\sin10^{\circ})=\frac{3}{4}.$$

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We need $$4\cos^220^\circ-4\cos20^\circ\sin10^\circ+4\sin^210^\circ=3$$

$$\iff2(1+\cos40^\circ)-2(\sin30^\circ-\sin10^\circ)+2(1-\cos20^\circ)=3$$

$$\iff\cos40^\circ-\cos20^\circ=-\sin10^\circ$$

which is evident from Prosthaphaeresis Formulas

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The obvious solution

$\cos3(20^\circ)=4\cos^320^\circ-3\cos20^\circ$ and $\sin3(10^\circ)=3\sin10^\circ-4\sin^310^\circ$ and $\cos3(20^\circ)=\sin3(10^\circ)$

Alternatively, dividing both sides by $\cos20^\circ+\sin10^\circ$ and replacing $\cos20^\circ=1-2\sin^210^\circ$

with $\sin10^\circ=s,3s-4s^3=\sin3(10^\circ)=\dfrac12$

$$4(1-2s^2)^2-4(1-2s^2)s+4s^2=3$$

$$\iff16s^4+8s^3-12s^2-4s+1=0$$

$$\iff-4s\left(3s-4s^3-\dfrac12\right)-2\left(3s-4s^3-\dfrac12\right)=0$$

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