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.
 A: You know that $\sin^2+\cos^2=1$. As $\sin(24)=p$ you have $p^2 + \cos^2(24)=1$. Solve this equation for $\cos(24)$.
For the second part you can use some trigonometric theorems.
For example 
$$ \sin(\alpha) \cdot \sin(\beta) = -\frac{1}{2} (\cos(\alpha+\beta) -\cos(\alpha - \beta))$$
This gives you 
$$\sin(168^\circ) \cdot \sin(-78^\circ)=-\frac{1}{2} ( \cos(90^\circ)- \cos(246^\circ))$$
We know that $\cos(90^\circ)=0$ and that $\cos(x+180^\circ)=-\cos(x)$ hence we have 
$$-\frac{1}{2} \cos(66^\circ)=-\frac{1}{2}\cos(90^\circ-24^\circ)=-\frac{1}{2} \sin(24^\circ)$$
A: 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*}
A: 
My favorite method
$$\sin \theta = \cfrac{\text {opposite}}{\text{hypothenus}}=\cfrac p1,\quad \quad  \cos\theta = \cfrac{\text {adjacent}}{\text{hypothenus}}$$
What is "adjacent"? By Pthagoras', we have  $\quad \text{adjacent}^2+\text{opposite}^2=\text{hypothenus}^2$
so $$ \text{adjacent}^2 + p^2=1 \implies \text{adjacent}=\sqrt{1-p^2}$$
et voila! $$\cos\theta = \cfrac{\text {adjacent}}{\text{hypothenus}}=\cfrac{\sqrt{1-p^2}}{1}=\sqrt{1-p^2}$$
A: 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 $\sin(\theta-180^\circ)=-\sin(180^\circ-\theta)=-\sin\theta$
