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Question Number 110269 by Khanacademy last updated on 28/Aug/20

Commented by bemath last updated on 28/Aug/20

−(1/2)

$$−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by Khanacademy last updated on 28/Aug/20

?

$$? \\ $$

Commented by john santu last updated on 28/Aug/20

i think this question is ⇒ cos ((2π)/7).cos ((4π)/7).cos ((6π)/7)

$${i}\:{think}\:{this}\:{question}\:{is}\: \\ $$$$\Rightarrow\:\:\:\:\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{7}}.\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{7}}.\mathrm{cos}\:\frac{\mathrm{6}\pi}{\mathrm{7}} \\ $$

Commented by bemath last updated on 28/Aug/20

if cos ((2π)/7).cos ((4π)/7).cos ((6π)/7) = p 2psin ((2π)/7)= sin ((4π)/7).cos ((4π)/7).cos ((6π)/7) 4psin ((2π)/7)=sin ((8π)/7).cos ((6π)/7) 8psin ((2π)/7)=sin ((14π)/7)+sin ((2π)/7) [ sin ((14π)/7) = 0 ] ⇒p=(1/8)

$${if}\:\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{7}}.\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{7}}.\mathrm{cos}\:\frac{\mathrm{6}\pi}{\mathrm{7}}\:=\:{p} \\ $$$$\mathrm{2}{p}\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{7}}=\:\mathrm{sin}\:\frac{\mathrm{4}\pi}{\mathrm{7}}.\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{7}}.\mathrm{cos}\:\frac{\mathrm{6}\pi}{\mathrm{7}} \\ $$$$\mathrm{4}{p}\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{7}}=\mathrm{sin}\:\frac{\mathrm{8}\pi}{\mathrm{7}}.\mathrm{cos}\:\frac{\mathrm{6}\pi}{\mathrm{7}} \\ $$$$\mathrm{8}{p}\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{7}}=\mathrm{sin}\:\frac{\mathrm{14}\pi}{\mathrm{7}}+\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{7}} \\ $$$$\left[\:\mathrm{sin}\:\frac{\mathrm{14}\pi}{\mathrm{7}}\:=\:\mathrm{0}\:\right]\:\Rightarrow{p}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$

Answered by Dwaipayan Shikari last updated on 28/Aug/20

cos2θ+cos4θ+cos6θ θ=(π/7) 7θ=π (1/(2sinθ))(sin3θ−sinθ+sin5θ−sin3θ+sin7θ−sin5θ) (1/(2sinθ))(sin7θ−sinθ)=−((sinθ)/(2sinθ))=−(1/2) (sin7θ=0)

$${cos}\mathrm{2}\theta+{cos}\mathrm{4}\theta+{cos}\mathrm{6}\theta\:\:\:\:\:\:\:\:\theta=\frac{\pi}{\mathrm{7}}\:\:\mathrm{7}\theta=\pi \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{sin}\theta}\left({sin}\mathrm{3}\theta−{sin}\theta+{sin}\mathrm{5}\theta−{sin}\mathrm{3}\theta+{sin}\mathrm{7}\theta−{sin}\mathrm{5}\theta\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{sin}\theta}\left({sin}\mathrm{7}\theta−{sin}\theta\right)=−\frac{{sin}\theta}{\mathrm{2}{sin}\theta}=−\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\left({sin}\mathrm{7}\theta=\mathrm{0}\right) \\ $$

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