<s>[INST] Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$ Please help me to prove this trigonometric equation.
$\cos \left( 36^\circ \right)-\cos \left( 72^\circ \right) = \frac{1}{2}$
Thank you.
 [/INST] As $\displaystyle\cos(180^\circ-y)=-\cos y$
$$S=\cos36^\circ-\cos72^\circ=\cos36^\circ+\cos(180^\circ-108^\circ)=\cos36^\circ+\cos108^\circ$$
Now multiplying the numerator & the denominator by $\displaystyle2\sin\frac{(108-36)}2^\circ,$
$$S=\frac{2\sin36^\circ\cos36^\circ+2\sin36^\circ\cos108^\circ}{2\sin36^\circ}$$
Using Werner Formula & $\sin2x$ formula,
$$S=\frac{\sin72^\circ+\sin144^\circ-\sin72^\circ}{2\sin36^\circ}=\frac{\sin(180^\circ-36^\circ)}{2\sin36^\circ}$$
Hope you can take it home from here
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