Question

If $\sin {18}^o=\frac{\sqrt{5}-1}{4}$, then what is the value of $\sin {81}^o$?

• A. $\frac{\sqrt{3+\sqrt{5}}+\sqrt{5-\sqrt{5}}}{4}$
• B. $\frac{\sqrt{3+\sqrt{5}}+\sqrt{5+\sqrt{5}}}{4}$
• C. $\frac{\sqrt{3-\sqrt{5}}+\sqrt{5-\sqrt{5}}}{4}$
• D. $\frac{\sqrt{3+\sqrt{5}}-\sqrt{5-\sqrt{5}}}{4}$

Answer 1

The correct option is A $\frac{\sqrt{3+\sqrt{5}}+\sqrt{5-\sqrt{5}}}{4}$

${\sin }^2\theta +{\cos }^2\theta =1$

$\therefore {\cos }^2\theta =1-{\sin }^2\theta$

$\therefore {\cos }^2\left({18}^0\right)=1-{\sin }^2\left({18}^0\right)$

$=1-(\frac{\sqrt{5}-1}{4}{)}^2$

$=1-\frac{5+1-2\sqrt{5}}{16}$

$=1-\frac{6-2\sqrt{5}}{16}$

$=\frac{16-(6-2\sqrt{5})}{16}$

$=\frac{10+2\sqrt{5}}{16}$

$\therefore \cos \left({18}^0\right)=\sqrt{\frac{10+2\sqrt{5}}{16}}$ ...(1)

$1+\cos 2\theta =2{\cos }^2\theta$

$\therefore \cos \theta =\sqrt{\frac{1+\cos 2\theta }{2}}$

$\therefore \cos 9^0=\sqrt{\frac{1+\cos {18}^0}{2}}$

$=\sqrt{\frac{1+\sqrt{\frac{10+2\sqrt{5}}{16}}}{2}}$ ...(from 1)

$=\sqrt{\frac{4+\sqrt{10+2\sqrt{5}}}{8}}$

$=\sqrt{\frac{8+2\sqrt{10+2\sqrt{5}}}{16}}$

$=\sqrt{\frac{3+\sqrt{5}+5-\sqrt{5}+2\sqrt{3+\sqrt{5}}\sqrt{5-\sqrt{5}}}{16}}$

$=\sqrt{\frac{(\sqrt{3+\sqrt{5}}{)}^2+(\sqrt{5-\sqrt{5}}{)}^2}{16}}$

$=\frac{\sqrt{3+\sqrt{5}}+\sqrt{5-\sqrt{5}}}{4}$

we know that $cos{9}^0=sin{81}^0$

So, the answer is option (A)