if $\sin24^\circ = p$ what is $\cos24^\circ$? Let $p=\sin 24^\circ$


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*Then what would $\cos (24^\circ)$ be in terms of $p$?

*What would $\sin (168^\circ) \cdot  \sin(-78^\circ)$ be in terms of $p$?
I'm not sure how to approach these as we have only started this section. 
Thanks.

Part 2
Method 1:
$\sin168^\circ=\sin(180^\circ-168^\circ)$ ( as $\sin(180^\circ-\theta)=\sin\theta$)
So, $\sin168^\circ=\sin12^\circ$
and $\sin(-78^\circ)=-\sin78^\circ$ as $\sin(-\theta)=-\sin\theta$
So, $\sin(-78^\circ)=-\sin(90^\circ-12^\circ)=-\cos12^\circ$
$$\implies \sin168^\circ\cdot\sin(-78^\circ)=\sin12^\circ\cdot(-\cos12^\circ)=-\frac{\sin (2\cdot12^\circ)}2=-\frac{\sin24^\circ}2$$
Method 2:
$\sin168^\circ=\sin(90^\circ+78^\circ)=\cos78^\circ$ as $\sin(90^\circ+\theta)=\cos\theta$
Again, $\cos78^\circ=\cos(-78^\circ)$ as $\cos(-\theta)=\cos\theta$
$$\implies \sin168^\circ\cdot\sin(-78^\circ)=\cos(-78^\circ)\cdot\sin(-78^\circ)=\frac{\sin 2(-78^\circ)}2$$
$$=\frac{\sin(24^\circ-180^\circ)}2=-\frac{\sin24^\circ}2$$ as $\sin(\theta-180^\circ)=-\sin(180^\circ-\theta)=-\sin\theta$
