Evaluating $\cos24^\circ-\cos84^\circ-\cos12^\circ+\sin42^\circ$

Question

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:

$$\cos24^\circ-\cos(60^\circ+24^\circ)-\cos12^\circ+\sin (12^\circ+30^\circ)$$

Answer 1

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

$\sin (54°)-\sin (18°)$

Answer 2

$$\begin{split} \cos24^\circ-\cos84^\circ-\cos12^\circ+\sin42^\circ &= \cos24^\circ-\cos84^\circ-\cos12^\circ+\cos48^\circ\\ &= \cos24^\circ +\cos48^\circ-\cos84^\circ-\cos12^\circ \\ &= 2\cos36^\circ\cos12^\circ-2\cos48^\circ\cos36^\circ\\ &=2\cos36^\circ(\cos12^\circ-\cos48^\circ)\\ &=2\cos36^\circ(2\sin18^\circ\sin30^\circ)\\ &=\frac{(2\sin18^\circ\cos18^\circ)2\cos36^\circ\sin30^\circ}{\cos18^\circ}\\ &= \frac{(2\sin36^\circ\cos36^\circ)\sin30^\circ}{\cos18^\circ}\\ &=\frac{\sin72^\circ\sin30^\circ}{\cos18^\circ}\\ &=\sin30^\circ=\frac{1}{2} \end{split} $$

Answer 3

$$-\cos84^\circ-\cos96^\circ$$

$$-\cos12^\circ=\cos192^\circ=\cos168^\circ$$

$$\sin42^\circ=\cos(?)$$

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 $5x=360^\circ n+120^\circ\iff x=72^\circ n+24^\circ; -2\le n\le2$

$$\sum_{n=-2}^2\cos(72^\circ n+24^\circ)=\dfrac0{16}$$

$$\implies\cos(-120)^\circ+\cos(-48^\circ)+\cos24^\circ+\cos96^\circ+\cos168^\circ=0$$