If-2pi-7-then-what-is-the-value-of-sin-sin2-sin4-

Question Number 99620 by Dwaipayan Shikari last updated on 22/Jun/20

If α=((2π)/7) then what is the value of (sinαsin2αsin4α)

$${If}\:\:\alpha=\frac{\mathrm{2}\pi}{\mathrm{7}}\:\:{then}\:{what}\:{is}\:{the}\:{value}\:{of}\:\left({sin}\alpha{sin}\mathrm{2}\alpha{sin}\mathrm{4}\alpha\right) \\ $$

Commented by Dwaipayan Shikari last updated on 22/Jun/20

I have found a way to solve it Suppose cosα+cos2α+cos4α=(1/(2sinα))[sin2α+sin3α−sinα+sin5α−sin3α] =(1/(2sinα))[sin((4π)/7)+sin((10π)/7)−sin((2π)/7)] =(1/(2sin((2π)/7)))[−sin((2π)/7)] =−(1/2) We can alsofind (sinα+sin2α+sin4α)=((√7)/2) So (sinαsin2αsin4α)=−(1/4)(sinα+sin2α+sin4α)=−((√7)/8) Please check this(Sir ,my process takes a long time and so large in size) So i can′t include every detailed prove

$${I}\:{have}\:{found}\:{a}\:{way}\:{to}\:{solve}\:{it} \\ $$$${Suppose} \\ $$$${cos}\alpha+{cos}\mathrm{2}\alpha+{cos}\mathrm{4}\alpha=\frac{\mathrm{1}}{\mathrm{2}{sin}\alpha}\left[{sin}\mathrm{2}\alpha+{sin}\mathrm{3}\alpha−{sin}\alpha+{sin}\mathrm{5}\alpha−{sin}\mathrm{3}\alpha\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}{sin}\alpha}\left[{sin}\frac{\mathrm{4}\pi}{\mathrm{7}}+{sin}\frac{\mathrm{10}\pi}{\mathrm{7}}−{sin}\frac{\mathrm{2}\pi}{\mathrm{7}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}{sin}\frac{\mathrm{2}\pi}{\mathrm{7}}}\left[−{sin}\frac{\mathrm{2}\pi}{\mathrm{7}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${We}\:{can}\:{alsofind}\:\left({sin}\alpha+{sin}\mathrm{2}\alpha+{sin}\mathrm{4}\alpha\right)=\frac{\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$$${So}\:\left({sin}\alpha{sin}\mathrm{2}\alpha{sin}\mathrm{4}\alpha\right)=−\frac{\mathrm{1}}{\mathrm{4}}\left({sin}\alpha+{sin}\mathrm{2}\alpha+{sin}\mathrm{4}\alpha\right)=−\frac{\sqrt{\mathrm{7}}}{\mathrm{8}} \\ $$$$ \\ $$$$ \\ $$$${Please}\:{check}\:{this}\left({Sir}\:,{my}\:{process}\:{takes}\:{a}\:{long}\:{time}\:{and}\:{so}\:{large}\:{in}\:{size}\right) \\ $$$$ \\ $$$${So}\:{i}\:{can}'{t}\:{include}\:{every}\:{detailed}\:{prove} \\ $$

Answered by MWSuSon last updated on 22/Jun/20

sin(((2π)/7))sin(((4π)/7))sin(((8π)/7))=−((√7)/8)

$$\mathrm{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\mathrm{sin}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)\mathrm{sin}\left(\frac{\mathrm{8}\pi}{\mathrm{7}}\right)=−\frac{\sqrt{\mathrm{7}}}{\mathrm{8}} \\ $$

Commented by Dwaipayan Shikari last updated on 22/Jun/20

$${Can}\:{you}\:{describe}\:{it}\:{sir}? \\ $$

Commented by MWSuSon last updated on 22/Jun/20

sir, do you mean the trig identity I used?

Commented by Dwaipayan Shikari last updated on 22/Jun/20

Sir I am student .so I want to know the process for solving it

Commented by MWSuSon last updated on 22/Jun/20

okay sir, I am also a student, to save time I just plugged in the value of the angle into the expression, but you can make use of the factor formulae to solve it.

Commented by john santu last updated on 22/Jun/20

$$\mathrm{wrong}.\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\frac{\sqrt{\mathrm{7}}}{\mathrm{8}} \\ $$

Commented by john santu last updated on 22/Jun/20

Commented by MWSuSon last updated on 22/Jun/20

sir I think there should be a negative sign. please recheck.

Commented by john santu last updated on 22/Jun/20

no. sin (π/7) >0 sin ((4π)/7) > 0 sin ((8π)/7) = sin (π+(π/7)) = sin (π/7) > 0

$$\mathrm{no}.\:\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\:>\mathrm{0} \\ $$$$\mathrm{sin}\:\frac{\mathrm{4}\pi}{\mathrm{7}}\:>\:\mathrm{0} \\ $$$$\mathrm{sin}\:\frac{\mathrm{8}\pi}{\mathrm{7}}\:=\:\mathrm{sin}\:\left(\pi+\frac{\pi}{\mathrm{7}}\right)\:=\:\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\:>\:\mathrm{0} \\ $$$$ \\ $$

Commented by bemath last updated on 22/Jun/20