If sin(frac{pi}{18}) sin(frac{5pi}{18}) sin(f | vedclass

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Question

(A) We use the trigonometric identity $\sin 	heta \sin(60^\circ - 	heta) \sin(60^\circ + 	heta) = \frac{1}{4} \sin 3	heta$.Here, $	heta = 10^\cir

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

Vedclass · Answer

(A) We use the trigonometric identity $\sin 	heta \sin(60^\circ - 	heta) \sin(60^\circ + 	heta) = \frac{1}{4} \sin 3	heta$.Here, $	heta = 10^\circ = \frac{\pi}{18}$.Then, $K = \sin 10^\circ \sin 50^\circ \sin 70^\circ = \frac{1}{4} \sin(3 	imes 10^\circ) = \frac{1}{4} \sin 30^\circ = \frac{1}{4} 	imes \frac{1}{2} = \frac{1}{8}$.Now, we need to find the value of $\sin(\frac{10K\pi}{3})$.Substituting $K = \frac{1}{8}$, we get $\sin(\frac{10 	imes (1/8) 	imes \pi}{3}) = \sin(\frac{10\pi}{24}) = \sin(\frac{5\pi}{12})$.Since $\frac{5\pi}{12} = 75^\circ$, we have $\sin(75^\circ) = \sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ$.$= (\frac{1}{\sqrt{2}} 	imes \frac{\sqrt{3}}{2}) + (\frac{1}{\sqrt{2}} 	imes \frac{1}{2}) = \frac{\sqrt{3}+1}{2\sqrt{2}}$.

If $\sin(\frac{\pi}{18}) \sin(\frac{5\pi}{18}) \sin(\frac{7\pi}{18}) = K$, then the value of $\sin(\frac{10K\pi}{3})$ is :

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