Answer
$\cos{18^o}=\dfrac{\sqrt{10+2\sqrt5}}{4}$
Work Step by Step
RECALL:
$\sin^2{x}+\cos^2{x}=1$
Solving for cosine in the identity above gives:
\begin{align*}
\cos^2{x}&=1-\sin^2{x}\\
\cos{x}&=\pm \sqrt{1-x^2}
\end{align*}
Use the identity above and the fact that $18^o$ is in quadrant I where cosine is positive to obtain:
\begin{align*}
\cos{18^o}&=\sqrt{1-\sin^2{18^o}}\\\\
&=\sqrt{1-\left(\frac{\sqrt5-1}{4}\right)^2}\\\\
&=\sqrt{1-\left(\frac{5-2\sqrt5+1}{16}\right)}\\\\
&=\sqrt{\frac{16}{16}-\left(\frac{6-2\sqrt5}{16}\right)}\\\\
&=\sqrt{\frac{16-6+2\sqrt5}{16}}\\\\
&=\sqrt{\frac{10+2\sqrt5}{16}}\\\\
&=\frac{\sqrt{10+2\sqrt5}}{4}\approx 0.9510565163\\\\
\end{align*}
Checking using a claculator gives:
$\cos{18^o}\approx 0.9510565163$
