Please prove it fast

1.	Please prove it fast
Answer» We know that, $\quad$ $\cos 36^{\circ}=\dfrac {1+\sqrt 5}{4}$ $\quad$ But we also know that, $\quad$ $\cos (90^{\circ}-x)=\sin x$ $\quad$ Thus, $\quad$ $\sin 54^{\circ}=\dfrac {1+\sqrt 5}{4}\\\\\\\sqrt {1-\cos^254^{\circ}}=\dfrac {1+\sqrt 5}{4}\\\\\\\cos 54^{\circ}=\dfrac {\sqrt{10-2\sqrt5}}{4}$ $\quad$ We know the identity, $\quad$ $\sin\left(\dfrac {x}{2}\right)=\sqrt {\dfrac {1-\cos x}{2}}$ $\quad$ Then, $\quad$ $\sin 27^{\circ}=\sqrt {\dfrac {1-\dfrac {\sqrt{10-2\sqrt5}}{4}}{2}}\\\\\\\sin 27^{\circ}=\sqrt {\dfrac {4-\sqrt{10-2\sqrt5}}{8}}\\\\\\\sin 27^{\circ}=\dfrac {\sqrt{8-2\sqrt{10-2\sqrt5}}}{4}\\\\\\4\sin 27^{\circ}=\sqrt{5+\sqrt5+3-\sqrt 5-2\sqrt{(5+\sqrt5)(3-\sqrt 5)}}\\\\\\4\sin 27^{\circ}=\sqrt{\left (\sqrt {5+\sqrt5}\right)^2+\left (\sqrt{3-\sqrt 5}\right)^2-2\sqrt{5+\sqrt5}\cdot\sqrt{3-\sqrt 5}}\\\\\\\underline {\underline {4\sin 27^{\circ}=\sqrt {5+\sqrt5}-\sqrt{3-\sqrt 5}}}$ $\quad$ Hence Proved! $\quad$ ====================================================== $\quad$ To obtain the value of $\cos 36^{\circ}$ we have to CONSIDER the golden triangle, given in the figure. $\quad$ Here ∆ABC is known as golden triangle because the sides are in golden ratio, i.e., $\quad$ $AC:AB:BC=1:\dfrac {1+\sqrt5}{2}:\dfrac {1+\sqrt5}{2}$ $\quad$ Or, according to the figure, $\quad$ $AC:AB:BC=\dfrac {\sqrt5-1}{2}:1:1$ $\quad$ One FEATURE about golden triangle is that the angle bisector of any of the 72° angle FORMS another golden triangle. ∆ADC is such one. $\quad$ So by the similarity of TRIANGLES, from ∆ABC and ∆ADC, we can obtain the relation, $\quad$ $X=\dfrac {1-X}{X}$ $\quad$ or, $\quad$ $X^2+X-1=0$ $\quad$ On solving this we get, $\quad$ $X=\dfrac {\sqrt 5-1}{2}$ $\quad$ On drawing an altitude from $\angle ADB$ to AB, we get the value of $\cos 36^{\circ}.$

Answer»

We know that,

$\quad$

$\cos 36^{\circ}=\dfrac {1+\sqrt 5}{4}$

$\quad$

But we also know that,

$\quad$

$\cos (90^{\circ}-x)=\sin x$

$\quad$

Thus,

$\quad$

$\sin 54^{\circ}=\dfrac {1+\sqrt 5}{4}\\\\\\\sqrt {1-\cos^254^{\circ}}=\dfrac {1+\sqrt 5}{4}\\\\\\\cos 54^{\circ}=\dfrac {\sqrt{10-2\sqrt5}}{4}$

$\quad$

We know the identity,

$\quad$

$\sin\left(\dfrac {x}{2}\right)=\sqrt {\dfrac {1-\cos x}{2}}$

$\quad$

Then,

$\quad$

$\sin 27^{\circ}=\sqrt {\dfrac {1-\dfrac {\sqrt{10-2\sqrt5}}{4}}{2}}\\\\\\\sin 27^{\circ}=\sqrt {\dfrac {4-\sqrt{10-2\sqrt5}}{8}}\\\\\\\sin 27^{\circ}=\dfrac {\sqrt{8-2\sqrt{10-2\sqrt5}}}{4}\\\\\\4\sin 27^{\circ}=\sqrt{5+\sqrt5+3-\sqrt 5-2\sqrt{(5+\sqrt5)(3-\sqrt 5)}}\\\\\\4\sin 27^{\circ}=\sqrt{\left (\sqrt {5+\sqrt5}\right)^2+\left (\sqrt{3-\sqrt 5}\right)^2-2\sqrt{5+\sqrt5}\cdot\sqrt{3-\sqrt 5}}\\\\\\\underline {\underline {4\sin 27^{\circ}=\sqrt {5+\sqrt5}-\sqrt{3-\sqrt 5}}}$