sin ^3 10^{circ}+sin ^3 50^{circ}-sin ^3 70^{ | vedclass

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Question

(A) We use the identity $\sin 3	heta = 3\sin 	heta - 4\sin ^3 	heta$,which implies $\sin ^3 	heta = \frac{3\sin 	heta - \sin 3	heta}{4}$.Applyin

Vedclass · Accepted Answer

$\frac{-3}{8}$

Vedclass · Answer

(A) We use the identity $\sin 3	heta = 3\sin 	heta - 4\sin ^3 	heta$,which implies $\sin ^3 	heta = \frac{3\sin 	heta - \sin 3	heta}{4}$.Applying this to each term:$\sin ^3 10^{\circ} = \frac{3\sin 10^{\circ} - \sin 30^{\circ}}{4}$$\sin ^3 50^{\circ} = \frac{3\sin 50^{\circ} - \sin 150^{\circ}}{4}$$\sin ^3 70^{\circ} = \frac{3\sin 70^{\circ} - \sin 210^{\circ}}{4}$Substituting these into the expression:$\frac{1}{4} [3(\sin 10^{\circ} + \sin 50^{\circ} - \sin 70^{\circ}) - (\sin 30^{\circ} + \sin 150^{\circ} - \sin 210^{\circ})]$Using $\sin 30^{\circ} = \frac{1}{2}$,$\sin 150^{\circ} = \frac{1}{2}$,and $\sin 210^{\circ} = -\frac{1}{2}$:$\sin 30^{\circ} + \sin 150^{\circ} - \sin 210^{\circ} = \frac{1}{2} + \frac{1}{2} - (-\frac{1}{2}) = \frac{3}{2}$Using $\sin 50^{\circ} - \sin 70^{\circ} = 2\cos 60^{\circ}\sin(-10^{\circ}) = -\sin 10^{\circ}$:$\sin 10^{\circ} + (\sin 50^{\circ} - \sin 70^{\circ}) = \sin 10^{\circ} - \sin 10^{\circ} = 0$Thus,the expression becomes $\frac{1}{4} [3(0) - \frac{3}{2}] = -\frac{3}{8}$.

$\sin ^3 10^{\circ}+\sin ^3 50^{\circ}-\sin ^3 70^{\circ}=$

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