sin 36^circ sin 72^circ sin 108^circ sin 144^ | vedclass

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Question

(D) We know that $\sin 108^\circ = \sin(180^\circ - 72^\circ) = \sin 72^\circ$ and $\sin 144^\circ = \sin(180^\circ - 36^\circ) = \sin 36^\circ$.There

Vedclass · Accepted Answer

$5/16$

Vedclass · Answer

(D) We know that $\sin 108^\circ = \sin(180^\circ - 72^\circ) = \sin 72^\circ$ and $\sin 144^\circ = \sin(180^\circ - 36^\circ) = \sin 36^\circ$.Therefore,the expression becomes $\sin^2 36^\circ \sin^2 72^\circ = (\sin 36^\circ \sin 72^\circ)^2$.Using the values $\sin 36^\circ = \frac{\sqrt{10 - 2\sqrt{5}}}{4}$ and $\sin 72^\circ = \frac{\sqrt{10 + 2\sqrt{5}}}{4}$:$\sin 36^\circ \sin 72^\circ = \frac{\sqrt{(10 - 2\sqrt{5})(10 + 2\sqrt{5})}}{16} = \frac{\sqrt{100 - 20}}{16} = \frac{\sqrt{80}}{16} = \frac{4\sqrt{5}}{16} = \frac{\sqrt{5}}{4}$.Squaring this result,we get $(\frac{\sqrt{5}}{4})^2 = \frac{5}{16}$.

$\sin 36^\circ \sin 72^\circ \sin 108^\circ \sin 144^\circ = $

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