The value of 4 sin 5^circ sin 55^circ sin 65^ | vedclass

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(B) Let $E = 4 \sin 5^\circ \sin 55^\circ \sin 65^\circ$. Using the identity $2 \sin A \sin B = \cos(A - B) - \cos(A + B)$,we have: $E = 2 [2 \sin 55^

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$\frac{\sqrt{3} - 1}{2\sqrt{2}}$

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(B) Let $E = 4 \sin 5^\circ \sin 55^\circ \sin 65^\circ$. Using the identity $2 \sin A \sin B = \cos(A - B) - \cos(A + B)$,we have: $E = 2 [2 \sin 55^\circ \sin 5^\circ] \sin 65^\circ$ $E = 2 [\cos(55^\circ - 5^\circ) - \cos(55^\circ + 5^\circ)] \sin 65^\circ$ $E = 2 [\cos 50^\circ - \cos 60^\circ] \sin 65^\circ$ Since $\cos 60^\circ = \frac{1}{2}$,we get: $E = 2 [\cos 50^\circ - \frac{1}{2}] \sin 65^\circ$ $E = 2 \cos 50^\circ \sin 65^\circ - \sin 65^\circ$ Using $2 \cos A \sin B = \sin(A + B) - \sin(A - B)$: $E = [\sin(50^\circ + 65^\circ) - \sin(50^\circ - 65^\circ)] - \sin 65^\circ$ $E = \sin 115^\circ - \sin(-15^\circ) - \sin 65^\circ$ $E = \sin 115^\circ + \sin 15^\circ - \sin 65^\circ$ Since $\sin 115^\circ = \sin(180^\circ - 65^\circ) = \sin 65^\circ$,we have: $E = \sin 65^\circ + \sin 15^\circ - \sin 65^\circ = \sin 15^\circ$ Using $\sin 15^\circ = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ$: $E = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} - 1}{2\sqrt{2}}$

The value of $4 \sin 5^\circ \sin 55^\circ \sin 65^\circ$ is equal to

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