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Question-80219




Question Number 80219 by Power last updated on 01/Feb/20
Answered by som(math1967) last updated on 01/Feb/20
let(π/7)=θ   ∴cos2θ+cos4θ+cos6θ  =(1/(2sinθ))(2cos2θsinθ+2sinθcos4θ                                                +2sinθcos6θ)  =(1/(2sinθ))(sin3θ−sinθ+sin5θ−sin3θ                                                         sin7θ−sin5θ)  (1/(2sinθ))(sin7θ−sinθ)★  =−(1/(2sinθ))×sinθ=−(1/2) ans  ★sin7θ=sinπ=0
$${let}\frac{\pi}{\mathrm{7}}=\theta\: \\ $$$$\therefore{cos}\mathrm{2}\theta+{cos}\mathrm{4}\theta+{cos}\mathrm{6}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{sin}\theta}\left(\mathrm{2}{cos}\mathrm{2}\theta{sin}\theta+\mathrm{2}{sin}\theta{cos}\mathrm{4}\theta\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{2}{sin}\theta{cos}\mathrm{6}\theta\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{sin}\theta}\left({sin}\mathrm{3}\theta−{sin}\theta+{sin}\mathrm{5}\theta−{sin}\mathrm{3}\theta\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{sin}\mathrm{7}\theta−{sin}\mathrm{5}\theta\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{sin}\theta}\left({sin}\mathrm{7}\theta−{sin}\theta\right)\bigstar \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}{sin}\theta}×{sin}\theta=−\frac{\mathrm{1}}{\mathrm{2}}\:{ans} \\ $$$$\bigstar{sin}\mathrm{7}\theta={sin}\pi=\mathrm{0} \\ $$
Commented by peter frank last updated on 01/Feb/20
great
$${great} \\ $$
Commented by Power last updated on 02/Feb/20
thanks
$$\mathrm{thanks} \\ $$

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