Evaluating $\cos24^\circ-\cos84^\circ-\cos12^\circ+\sin42^\circ$ How to evaluate:
$$\cos24^\circ-\cos84^\circ-\cos12^\circ+\sin42^\circ$$
Can somebody help me handle it?
I have no idea what to do.
This is my attempt:
$$\cos24^\circ-\cos(60^\circ+24^\circ)-\cos12^\circ+\sin (12^\circ+30^\circ)$$
 A: Try proving $\sin (18°)=\frac {\sqrt{5}-1}4 $
Hint : If $\theta=18°$ Then $5\theta=90°$
So $2\theta=90°-3\theta$
$\Rightarrow \sin(2\theta)=\cos (3\theta) $   and so on.
With the help of $\sin (18°) $, find $\sin (54°) $
Now ,
$\cos 24-\cos 84=2\sin 54 \sin 30$
$\cos 12-\sin 42=\cos 12-\cos 48$
$\quad \quad \quad \quad =2\sin (30) \sin (18)$
So the required value is
$\sin (54°)-\sin (18°)$
A: $$\begin{split}
\cos24^\circ-\cos84^\circ-\cos12^\circ+\sin42^\circ  &= \cos24^\circ-\cos84^\circ-\cos12^\circ+\cos48^\circ\\
&= \cos24^\circ +\cos48^\circ-\cos84^\circ-\cos12^\circ \\
&= 2\cos36^\circ\cos12^\circ-2\cos48^\circ\cos36^\circ\\
&=2\cos36^\circ(\cos12^\circ-\cos48^\circ)\\
&=2\cos36^\circ(2\sin18^\circ\sin30^\circ)\\
&=\frac{(2\sin18^\circ\cos18^\circ)2\cos36^\circ\sin30^\circ}{\cos18^\circ}\\
&= \frac{(2\sin36^\circ\cos36^\circ)\sin30^\circ}{\cos18^\circ}\\
&=\frac{\sin72^\circ\sin30^\circ}{\cos18^\circ}\\
&=\sin30^\circ=\frac{1}{2}
\end{split}
$$
A: $$-\cos84^\circ-\cos96^\circ$$
$$-\cos12^\circ=\cos192^\circ=\cos168^\circ$$
$$\sin42^\circ=\cos(?)$$
As $\cos5x=-\dfrac12$ for  $x=120^\circ,240^\circ\equiv-120^\circ,480^\circ\equiv120^\circ,960^\circ\equiv-120^\circ\pmod{360^\circ}$
Now $\cos5x+\cos x=2\cos3x\cos2x$
$\cos5x=2(4\cos^3x-3\cos x)(2\cos^2x-1)=?$
So, the roots of $$16c^5-20c^3+16c+\dfrac12=0$$ are $5x=360^\circ n+120^\circ\iff x=72^\circ n+24^\circ; -2\le n\le2$
$$\sum_{n=-2}^2\cos(72^\circ n+24^\circ)=\dfrac0{16}$$
$$\implies\cos(-120)^\circ+\cos(-48^\circ)+\cos24^\circ+\cos96^\circ+\cos168^\circ=0$$
