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Question Number 108644 by bemath last updated on 18/Aug/20

$$\frac{bemath}{★}$$$$findthevalueof$$$$\sin \left(\frac{\pi }{9}\right)\sin \left(\frac{2\pi }{9}\right)\sin \left(\frac{3\pi }{9}\right)\sin \left(\frac{4\pi }{9}\right)?$$

Commented by bemath last updated on 18/Aug/20

$$greatanswerallmaster$$

Answered by mnjuly1970 last updated on 18/Aug/20

$$4\sin \left(\mathrm{x}\right)\sin (60-\mathrm{x})\sin (60+\mathrm{x})\stackrel{⟨\mathrm{identity}⟩}{=}\sin \left(3\mathrm{x}\right)$$$$\mathrm{x}={20}^{°}\Rightarrow 4\sin \left(20\right)\sin \left(40\right)\sin \left(80\right)=\sin \left(60\right)=\frac{\sqrt{3}}{2}$$$$\sin \left(60\right)\{\sin \left(20\right)\sin \left(40\right)\sin \left(80\right)=\frac{\sqrt{3}}{8}\}$$$$\sin \left(20\right)\sin \left(40\right)\sin \left(60\right)\sin \left(80\right)=\frac{3}{16}..$$$$$$

Answered by Dwaipayan Shikari last updated on 18/Aug/20

$$sin\theta sin2\theta sin3\theta sin4\theta (\frac{\pi }{9}=\theta \pi +\theta =10\theta \Rightarrow -cos\theta =cos10\theta )$$$$(\pi =9\theta -cos3\theta =cos6\theta ,-cos\theta =cos8\theta$$$$=\frac{1}{4}(cos3\theta -cos5\theta )(cos\theta -cos5\theta )$$$$=\frac{1}{4}(cos3\theta cos\theta -cos5\theta cos\theta -cos3\theta cos5\theta +{cos}^{2}5\theta )$$$$=\frac{1}{4}(\frac{1}{2}cos4\theta +\frac{1}{2}cos2\theta -\frac{1}{2}cos6\theta -\frac{1}{2}cos4\theta -\frac{1}{2}cos8\theta -\frac{1}{2}cos2\theta +\frac{1}{2}(1+cos10\theta )$$$$=\frac{1}{4}(-\frac{1}{2}cos6\theta -\frac{1}{2}cos8\theta +\frac{1}{2}-\frac{1}{2}cos\theta )$$$$=\frac{1}{4}(\frac{1}{2}cos3\theta +\frac{1}{2})$$$$=\frac{1}{4}(\frac{1}{4}+\frac{1}{2})$$$$=\frac{3}{16}$$

Answered by nimnim last updated on 18/Aug/20

$$=sin20sin40sin60sin80$$$$=\frac{1}{2}(cos20-cos60)\frac{\sqrt{3}}{2}sin80$$$$=\frac{\sqrt{3}}{4}sin80cos20-\frac{\sqrt{3}}{4}\times \frac{1}{2}sin80$$$$=\frac{\sqrt{3}}{8}(sin100-sin60)-\frac{\sqrt{3}}{8}sin80$$$$=\frac{\sqrt{3}}{8}(sin100-sin80)+\frac{\sqrt{3}}{8}sin60$$$$=\frac{\sqrt{3}}{8}\left(2cos90sin10\right)+\frac{\sqrt{3}}{8}\times \frac{\sqrt{3}}{2}$$$$=0+\frac{3}{16}\{\because cos90=0\}$$$$=\frac{3}{16}★$$

Answered by bobhans last updated on 18/Aug/20

$$\frac{\mathbb{B}ob\mathbb{H}ans}{⋮\dots ⋮}$$$$let\sin \left(\frac{\pi }{9}\right)\sin \left(\frac{2\pi }{9}\right)\sin \left(\frac{3\pi }{9}\right)\sin \left(\frac{4\pi }{9}\right)=r$$$$\Rightarrow r=\frac{\sqrt{3}}{2}\sin \left(20°\right)\sin \left(40°\right)\cos \left(10°\right)$$$$\Rightarrow r=\frac{\sqrt{3}}{4}\sin \left(20°\right)[\sin \left(50°\right)+\sin \left(30°\right)]$$$$\Rightarrow r=\frac{\sqrt{3}}{8}\sin \left(20°\right)+\frac{\sqrt{3}}{4}\sin \left(50°\right)\sin \left(20°\right)$$$$\Rightarrow r=\frac{\sqrt{3}}{8}\sin \left(20°\right)-\frac{\sqrt{3}}{8}\{-2\sin \left(50°\right)\sin \left(20°\right)\}$$$$\Rightarrow r=\frac{\sqrt{3}}{8}\sin \left(20°\right)-\frac{\sqrt{3}}{8}(\cos \left(70°\right)-\cos \left(30°\right)\}$$$$\Rightarrow r=\frac{\sqrt{3}}{8}\sin \left(20°\right)-\frac{\sqrt{3}}{8}\sin \left(20°\right)+\frac{\sqrt{3}}{8}.\frac{\sqrt{3}}{2}$$$$\Rightarrow r=\frac{3}{16}$$

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