Question

please prove it fast ​

Answer 1

We know that,

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

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But we also know that,

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

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Thus,

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

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We know the identity,

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

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Then,

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

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Hence Proved!

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To obtain the value of $\cos 36^{\circ}$ we have to consider the golden triangle, given in the figure.

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Here ∆ABC is known as golden triangle because the sides are in golden ratio, i.e.,

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$AC:AB:BC=1:\dfrac {1+\sqrt5}{2}:\dfrac {1+\sqrt5}{2}$

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Or, according to the figure,

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$AC:AB:BC=\dfrac {\sqrt5-1}{2}:1:1$

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One feature about golden triangle is that the angle bisector of any of the 72° angle forms another golden triangle. ∆ADC is such one.

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So by the similarity of triangles, from ∆ABC and ∆ADC, we can obtain the relation,

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$X=\dfrac {1-X}{X}$

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or,

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$X^2+X-1=0$

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On solving this we get,

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$X=\dfrac {\sqrt 5-1}{2}$

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On drawing an altitude from $\angle ADB$ to AB, we get the value of $\cos 36^{\circ}.$