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Circular Measurement of Angle

cos18o - sin1...

Question

$cos 18^{o} - sin 18^{o} =$

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Solution

$c o s 18^{\circ} - s i n 18^{\circ}$
$\Rightarrow \sqrt{\frac{5 + \sqrt{5}}{2}} - (\frac{\sqrt{5} - 1}{4})$

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Q. Prove that :

cos 18^{o} - sin 18^{o} = \sqrt{2} sin 27^{o}

c o s 18^{o}

\frac{{c o s 18}^{o}}{{s i n 72}^{o}}

cos 72^{o} - sin 18^{o} =

\frac{sin 18^{o}}{cos 72^{o}} =

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