Question

What is the value of cos ${36}^o$ ?

• A. $\frac{\sqrt{5}-1}{4}$
• B. $\frac{\sqrt{5}+1}{4}$
• C. $\frac{\sqrt{10+2\sqrt{5}}}{4}$
• D. $\frac{\sqrt{10-2\sqrt{5}}}{4}$

Answer 1

Answer: (B)

$\frac{\sqrt{5}+1}{4}$

Let $A={18}^o$

$\Rightarrow 5A={90}^o\Rightarrow 2A+3A=90˚\Rightarrow 2A={90}^o-3A$

Taking $\sin$ on both sides, we get

$\sin 2A=\sin ({90}^o-3A)=\cos 3A$

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

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

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

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

$\Rightarrow 4{\sin }^{2}A+2\sin A-1=0$,

which is a quadratic in $\sin A$

Therefore, $\sin A=\frac{-2\pm \sqrt{2^2-\mathrm{4.4.}(-1)}}{2.4}$

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

Now $\sin {18}^o$ is positive, as ${18}^o$ lies in first quadrant.

Therefore, $\sin {18}^o=\sin A=\frac{-1\pm \sqrt{5}}{4}$

Now, $\cos {36}^o=\cos (2\times 18°)$

$\Rightarrow \cos {36}^o=1-2{\sin }^2{18}^o$

$\Rightarrow \cos {36}^o=1-2{\left(\frac{\sqrt{5}-1}{4}\right)}^2$

Therefore, $\cos {36}^o=\frac{\sqrt{5}+1}{4}$