Trigonometric Bases

Let the Brilliantica's hypotenuse $AB=1$ , then $\cos_{60^\circ} x = AD$ and $\cos_{120^\circ} x = AC$ . We note that $\triangle BCD$ is equilateral. Since $BE = \sin x$ , $BC=CD=DB=\frac 2{\sqrt 3}\sin x$ and $CE=ED = \frac 1{\sqrt 3}\sin x$ . Then we have:

$\begin{cases} \cos_{60^\circ} x = AD = AE+ED = \cos x + \dfrac {\sin x}{\sqrt 3} \\ \cos_{120^\circ} x = AC = AE-CE = \cos x - \dfrac {\sin x}{\sqrt 3} \end{cases}$

Then $\color{#3D99F6} \cos_{60^\circ} x + \cos_{120^\circ} x = 2\cos x$ . Now we have:

$\begin{aligned} \cos^3_{60^\circ} x + \cos^3_{120^\circ} x & = \left(\cos x + \frac {\sin x}{\sqrt 3}\right)^3 + \left(\cos x - \frac {\sin x}{\sqrt 3}\right)^3 \\ & = 2 \cos^3 x + 2 \sin^2 x \cos x \\ & = 2\cos x \left(\cos^2 x + \sin^2 x\right) \\ & = \color{#3D99F6} 2\cos x \\ & = \boxed{\color{#3D99F6}\cos_{60^\circ} x + \cos_{120^\circ} x} \end{aligned}$

Let $\angle A = x$ , $\angle C = 60°$ , and $c = 1$ . Then $a = \sin_{60°} x$ and $b = \cos_{60°} x$ , and by law of cosines, $\sin_{60°}^2 x - \sin_{60°} x \cos_{60°} x + \cos_{60°}^2 x = 1$ , so $\cos_{60°}^3 x + \sin_{60°}^3 x = (\cos_{60°} x + \sin_{60°} x)(\sin_{60°}^2 x - \sin_{60°} x \cos_{60°} x + \cos_{60°}^2 x) = \cos_{60°} x + \sin_{60°} x$ .

Let $\angle A = x$ , $\angle C’ = 120°$ , and $c = 1$ . Then $a’ = \sin_{120°} x$ and $b’ = \cos_{120°} x$ , and by law of cosines, $\sin_{120°}^2 x + \sin_{120°} x \cos_{120°} x + \cos_{120°}^2 x = 1$ , so $\cos_{60°}^3 x - \sin_{60°}^3 x = (\cos_{60°} x - \sin_{60°} x)(\sin_{60°}^2 x + \sin_{60°} x \cos_{60°} x + \cos_{60°}^2 x) = \cos_{60°} x - \sin_{60°} x$ .

Also, $a$ and $a’$ are sides of an equilateral triangle, so $a = a’$ or $\sin_{60°} x = \sin_{120°} x$ .

Therefore,

$\cos_{60°}^3 x + \cos_{120°}^3 x$

$=$ $\cos_{60°}^3 x + \sin_{60°}^3 x + \cos_{120°}^3 x - \sin_{120°}^3 x$

$=$ $\cos_{60°} x + \sin_{60°} x + \cos_{120°} x - \sin_{120°} x$

$=$ $\boxed{\cos_{60°} x + \cos_{120°} x}$ .