## Trigonometry (11th Edition) Clone

$\cos{18^o}=\dfrac{\sqrt{10+2\sqrt5}}{4}$
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: $\cos{18^o}\approx 0.9510565163$