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Question Number 121256 by ZiYangLee last updated on 06/Nov/20

Prove 2sin^2 3θ−2sin^2 θ=cos2θ−cos6θ. By substitusing θ=(π/(10)) in the above identity, prove that sin((3π)/(10))−sin(π/(10))=(1/2)

$$\mathrm{Prove}\:\mathrm{2sin}^{\mathrm{2}} \mathrm{3}\theta−\mathrm{2sin}^{\mathrm{2}} \theta=\mathrm{cos2}\theta−\mathrm{cos6}\theta. \\ $$$$\mathrm{By}\:\mathrm{substitusing}\:\theta=\frac{\pi}{\mathrm{10}}\:\mathrm{in}\:\mathrm{the}\:\mathrm{above}\:\mathrm{identity}, \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{sin}\frac{\mathrm{3}\pi}{\mathrm{10}}−\mathrm{sin}\frac{\pi}{\mathrm{10}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Answered by TANMAY PANACEA last updated on 06/Nov/20

2sin^2 3θ−2sin^2 θ 1−cos6θ−1+cos2θ cos2θ−cos6θ so sin3θ−sinθ=((cos2θ−cos6θ)/(2(sin3θ+sinθ))) sin((3π)/(10))−sin(π/(10)) ((cos((2π)/(10))−cos((6π)/(10)))/(2(sin((3π)/(10))+sin(π/(10))))) (1/2)×((2sin(((4π)/(10)))sin(((2π)/(20))))/(2sin(((2π)/(10)))cos((π/(10)))))=(1/2) proved note sin((4π)/(10))=cos((π/2)−((4π)/(10)))=cos((π/(10)))

$$\mathrm{2}{sin}^{\mathrm{2}} \mathrm{3}\theta−\mathrm{2}{sin}^{\mathrm{2}} \theta \\ $$$$\mathrm{1}−{cos}\mathrm{6}\theta−\mathrm{1}+{cos}\mathrm{2}\theta \\ $$$${cos}\mathrm{2}\theta−{cos}\mathrm{6}\theta \\ $$$$\left.{so}\:{sin}\mathrm{3}\theta−{sin}\theta=\frac{{cos}\mathrm{2}\theta−{cos}\mathrm{6}\theta}{\mathrm{2}\left({sin}\mathrm{3}\theta+{sin}\theta\right.}\right) \\ $$$${sin}\frac{\mathrm{3}\pi}{\mathrm{10}}−{sin}\frac{\pi}{\mathrm{10}} \\ $$$$\frac{{cos}\frac{\mathrm{2}\pi}{\mathrm{10}}−{cos}\frac{\mathrm{6}\pi}{\mathrm{10}}}{\mathrm{2}\left({sin}\frac{\mathrm{3}\pi}{\mathrm{10}}+{sin}\frac{\pi}{\mathrm{10}}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}{sin}\left(\frac{\mathrm{4}\pi}{\mathrm{10}}\right){sin}\left(\frac{\mathrm{2}\pi}{\mathrm{20}}\right)}{\mathrm{2}{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{10}}\right){cos}\left(\frac{\pi}{\mathrm{10}}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\:{proved} \\ $$$${note} \\ $$$${sin}\frac{\mathrm{4}\pi}{\mathrm{10}}={cos}\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{4}\pi}{\mathrm{10}}\right)={cos}\left(\frac{\pi}{\mathrm{10}}\right) \\ $$

Commented by ZiYangLee last updated on 08/Nov/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$$$ \\ $$

Commented by TANMAY PANACEA last updated on 08/Nov/20

most welcome

$${most}\:{welcome} \\ $$

Answered by Dwaipayan Shikari last updated on 06/Nov/20

sin((3π)/(10))−sin(𝛑/(10)) =(((√5)+1)/4)−(((√5)−1)/4)=(1/2)

$${sin}\frac{\mathrm{3}\pi}{\mathrm{10}}−\boldsymbol{{sin}}\frac{\boldsymbol{\pi}}{\mathrm{10}} \\ $$$$=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}}−\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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