Question-148754

Question Number 148754 by mathdanisur last updated on 30/Jul/21

Commented by liberty last updated on 31/Jul/21

ans : 850

Answered by liberty last updated on 31/Jul/21

\frac{\sin^{6} 36 °}{\cos^{6} 36 °} + \frac{\sin^{6} 72 °}{\cos^{6} 72 °} =

\frac{\frac{1}{2^{6}} {(2 \cos 72 ° \sin 36 °)}^{6} + \frac{1}{2^{6}} {(2 \sin 72 ° \cos 36 °)}^{2}}{\frac{1}{2^{6}} {(2 \cos 72 ° \cos 36 °)}^{6}}

= \frac{{(\sin 108 ° - \sin 36 °)}^{6} + {(\sin 108 ° + \sin 36 °)}^{6}}{{(\sin 18 ° (1 - 2 \sin 18 °))}^{6}}

= \frac{\cos^{6} 18 ° {(1 - 2 \sin 18 °)}^{6} + \cos^{6} 18 ° {(1 + 2 \sin 18 °)}^{6}}{\sin^{6} 18 ° {(1 - 2 \sin 18 °)}^{6}}

you can finish it

Commented by liberty last updated on 31/Jul/21

\cos 18 ° = \frac{\sqrt{2 \sqrt{5} + 10}}{4}

\sin 18 ° = \frac{\sqrt{5} - 1}{4}

Commented by mathdanisur last updated on 31/Jul/21

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