Question

Show that $4\sin {27}^{\circ }={(5+\sqrt{5})}^{1/2}+{(3-\sqrt{5})}^{1/2}$.

Answer 1

lets take

${(\sin {27}^o+\cos {27}^o)}^2={\sin }^2{27}^o+{\cos }^2{27}^o+2\sin {27}^o\cos {27}^o=1+\sin {54}^o$(using identity $sin2\theta =2sin\theta cos\theta$)

$\Rightarrow {(\sin {27}^o+\cos {27}^o)}^2=1+\sin {54}^o=1+\frac{\sqrt{5}+1}{4}=\frac{1}{4}(5+\sqrt{5})$

$\Rightarrow \sin {27}^o+\cos {27}^o=\frac{1}{2}\sqrt{5+\sqrt{5}}$ ...(1)
Similarly
${(\sin {27}^o-\cos {27}^o)}^2=1-\cos {36}^o=1-\frac{\sqrt{5}+1}{4}$
$\Rightarrow \sin {27}^o-\cos {27}^o=\frac{1}{2}\sqrt{3-\sqrt{5}}$ ...(2)
From (1) and (2)
$4\sin {27}^o=\sqrt{5+\sqrt{5}}-\sqrt{3-\sqrt{5}}$