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

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


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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Trigonometry

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