Question

Find the value of $\cos {18}^{\circ }$

Answer 1

Find the value of $\cos 18°$ using multiple angle formulae of trigonometric ratios

Let angle $A=18°$

Therefore, $5A=90°$

$\Rightarrow 2A+3A=90˚$

$\Rightarrow$ $2A=90˚-3A$

Taking $\sin e$ on both sides, we get
$\sin 2A=\sin (90˚-3A)=\cos 3A$

$\Rightarrow$ $2\sin A\times \cos A=4{\cos }^{3}A-3\cos A$

$\Rightarrow 2\sin A\times \cos A-4{\cos }^{3}A+3\cos A=0$

$\Rightarrow$ $\cos A(2\sin A-4{\cos }^{2}A+3)=0$

Dividing both sides by$\cos A=\cos 18˚\ne 0$, we get

$2\sin A-4(1-{\sin }^{2}A)+3=0$

$\Rightarrow 4{\sin }^{2}A+2\sin A-1=0$, which is a quadratic equation in $\sin A$

$\Rightarrow \sin A=\frac{-2\pm \sqrt{{\left(2\right)}^2-4\left(4\right)\left(-1\right)}}{2\left(4\right)}$

$$\begin{gathered} =\frac{-2\pm \sqrt{16+4}}{8} \\ =\frac{-2\pm 2\sqrt{5}}{8} \end{gathered}$$

$\Rightarrow$$\sin A=\frac{-1\pm \sqrt{5}}{4}$

Now the value of $\cos {18}^{\circ }$ is

$\Rightarrow$$\cos {18}^{\circ }=\sqrt{1-{\sin }^2{18}^{\circ }}$

$$\begin{gathered} =\sqrt{1-{\left(\frac{\sqrt{5}-1}{4}\right)}^2} \\ =\sqrt{1-\left(\frac{5+1-2\sqrt{5}}{16}\right)} \\ =\sqrt{\frac{10+2\sqrt{5}}{16}} \end{gathered}$$

$\Rightarrow$$\cos {18}^{\circ }=\frac{\sqrt{10+2\sqrt{5}}}{4}$

Hence the value of $\cos {18}^{\circ }=\frac{\sqrt{10+2\sqrt{5}}}{4}$