# Show that $\cos^220^\circ-\cos20^\circ\sin10^\circ+\sin^210^\circ=\frac34$

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

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

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

• 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... – Dan Uznanski Feb 18 at 19:15

That last equation is trivial because $$\sin 3\theta=3\sin \theta-4\sin^3\theta$$.

• Thanks, as I made more edits I gradually came to the same conclusion :) – user170231 Feb 18 at 19:19

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 $$4(\cos^3 c^\circ+\sin^3 s^\circ)=3(\cos c^\circ+\sin s^\circ).$$

• @user170231 I propose a complex numbers solution and a generalization. – Robert Z Feb 18 at 19:53

We can use also the following way. $$\cos^220^\circ-\cos20^\circ\sin10^\circ+\sin^210^\circ=$$ $$=\frac{1}{2}(1+\cos40^{\circ})-\frac{1}{2}(\sin30^{\circ}-\sin10^{\circ})+\frac{1}{2}(1-\cos20^{\circ})=$$ $$=\frac{3}{4}+\frac{1}{2}(\cos40^{\circ}+\sin10^{\circ}-\cos20^{\circ})=$$ $$=\frac{3}{4}-\frac{1}{2}(\sin10^{\circ}-2\sin30^{\circ}\sin10^{\circ})=\frac{3}{4}.$$

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

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

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

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

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