Question

Solve and find the value of $\sin {36}^{\circ }$.

Answer 1

We know $\sin {18}^{\circ }=\frac{\sqrt{5}-1}{4},\cos {18}^{\circ }=\frac{\sqrt{10+2\sqrt{5}}}{4}$

Step 1: Finding the value of $\cos {36}^{\circ }$

Here we are going to use the formula $\cos 2\theta =1-2{\sin }^2\theta$

$$\begin{gathered} \cos {36}^{\circ }=\cos 2\left({18}^{\circ }\right)=1-2{\sin }^2\left({18}^{\circ }\right) \\ =1-2{\left[\frac{\sqrt{5}-1}{4}\right]}^2 \\ =1-\cancel{2}\left[\frac{5+1-2\sqrt{5}}{{\cancel{16}}_8}\right] \\ =1-\frac{\left(6-2\sqrt{5}\right)}{8} \\ =\frac{8-\left(6-2\sqrt{5}\right)}{8} \\ =\frac{8-6+2\sqrt{5}}{8} \\ =\frac{2+2\sqrt{5}}{8} \\ =\frac{2(1+\sqrt{5)}}{8} \\ =\frac{1+\sqrt{5}}{4} \\ \therefore \cos {36}^{\circ }=\frac{1+\sqrt{5}}{4} \end{gathered}$$

Step 2: Finding the value of $\sin {36}^{\circ }$

Here we will be using the formula ${\sin }^2\theta =1-{\cos }^2\theta$

$$\begin{gathered} \Rightarrow {\sin }^2{36}^{\circ }=1-{\cos }^2{36}^{\circ } \\ =1-{\left[\frac{\sqrt{5}+1}{4}\right]}^2 \\ =1-\left[\frac{5+1+2\sqrt{5}}{16}\right] \\ =\frac{16-(6+2\sqrt{5)}}{16} \\ =\frac{10-2\sqrt{5}}{16} \\ \therefore {\sin }^{2}36=\frac{10-2\sqrt{5}}{16} \\ \Rightarrow \sin 36=\sqrt{\frac{10-2\sqrt{5}}{16}} \\ \therefore \sin {36}^{{\circ }^{}}=\frac{\sqrt{10-2\sqrt{5}}}{4} \end{gathered}$$

Therefore, $\sin {36}^{\circ }=\frac{\sqrt{10-2\sqrt{5}}}{4}$.