<s>[INST] if $\sin24^\circ = p$ what is $\cos24^\circ$? Let $p=\sin 24^\circ$


*

*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.
 [/INST] Part 1
Use the identity
\begin{equation*}
\sin^2A+\cos^2A=1\\
\cos^2 A={1-\sin^2 A}
\end{equation*}
so
\begin{align*}
\cos^2 24^\circ&={1-\sin^2( 24^\circ) }\\
\\
\\cos^2 24^\circ&={1-p^2 }
\\
\\
\cos24^\circ&=\sqrt{1-p^2 }
\end{align*}
Here we only take positive value of square root because $24^\circ$ is in I quaderent and cos ratio in quaderent I is positive.
Part 2
\begin{align*}
\sin 168^\circ \cdot \sin(-78^\circ)
&=-\frac{2\sin 168^\circ\cdot\sin78^\circ}{2}
\end{align*}
Using the formula ${2\sin A\cdot\sin B}=\cos (A-B)-\cos(A+B)$
\begin{align*}
\sin(168^\circ)\cdot \sin(-78^\circ)&=-\frac{\cos 90^\circ-\cos 246^\circ}{2} \\
&=\frac{\cos 246^\circ}{2}\\
&=\frac{\cos(270^\circ-24^\circ)}{2}\\
&=-\frac{\sin 24^\circ}{2}\\
&=-\frac{p}{2}
\end{align*}
</s>