Question-115136

Question Number 115136 by Algoritm last updated on 23/Sep/20

Answered by Olaf last updated on 23/Sep/20

sin2θ = 2sinθcosθ cosθ = (1/2).((sin2θ)/(sinθ)) θ = ((kπ)/(2n+1)), cos(((kπ)/(2n+1))) = (1/2).((sin(((2kπ)/(2n+1))))/(sin(((kπ)/(2n+1))))) Π_(k=1) ^n cos(((kπ)/(2n+1))) = (1/2^n )Π_(k=1) ^n ((sin(((2kπ)/(2n+1))))/(sin(((kπ)/(2n+1))))) Π_(k=1) ^n cos(((kπ)/(2n+1))) = (1/2^n ).((sin(((2π)/(2n+1))))/(sin((π/(2n+1)))))×((sin(((4π)/(2n+1))))/(sin(((2π)/(2n+1)))))...×((sin(((2πn)/(2n+1))))/(sin(((πn)/(2n+1))))) Π_(k=1) ^n cos(((kπ)/(2n+1))) = (1/2^n ).((sin(((2πn)/(2n+1))))/(sin((π/(2n+1))))) sin(((2πn)/(2n+1))) = sin(π−(π/(2n+1))) = sin((π/(2n+1))) ⇒ Π_(k=1) ^n cos(((kπ)/(2n+1))) = (1/2^n )

$$\mathrm{sin2}\theta\:=\:\mathrm{2sin}\theta\mathrm{cos}\theta \\ $$$$\mathrm{cos}\theta\:=\:\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{sin2}\theta}{\mathrm{sin}\theta} \\ $$$$\theta\:=\:\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}},\:\mathrm{cos}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{sin}\left(\frac{\mathrm{2}{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)}{\mathrm{sin}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\mathrm{cos}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{\mathrm{sin}\left(\frac{\mathrm{2}{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)}{\mathrm{sin}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\mathrm{cos}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }.\frac{\mathrm{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)}{\mathrm{sin}\left(\frac{\pi}{\mathrm{2}{n}+\mathrm{1}}\right)}×\frac{\mathrm{sin}\left(\frac{\mathrm{4}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)}{\mathrm{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)}…×\frac{\mathrm{sin}\left(\frac{\mathrm{2}\pi{n}}{\mathrm{2}{n}+\mathrm{1}}\right)}{\mathrm{sin}\left(\frac{\pi{n}}{\mathrm{2}{n}+\mathrm{1}}\right)} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\mathrm{cos}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }.\frac{\mathrm{sin}\left(\frac{\mathrm{2}\pi{n}}{\mathrm{2}{n}+\mathrm{1}}\right)}{\mathrm{sin}\left(\frac{\pi}{\mathrm{2}{n}+\mathrm{1}}\right)} \\ $$$$\mathrm{sin}\left(\frac{\mathrm{2}\pi{n}}{\mathrm{2}{n}+\mathrm{1}}\right)\:=\:\mathrm{sin}\left(\pi−\frac{\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\:=\:\mathrm{sin}\left(\frac{\pi}{\mathrm{2}{n}+\mathrm{1}}\right) \\ $$$$\Rightarrow\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\mathrm{cos}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} } \\ $$$$ \\ $$$$ \\ $$

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