2 Answers

Answer 1 · 2017-06-11T08:05:35

$sin (\frac{2 π}{7}) + sin (\frac{4 π}{7}) + sin (\frac{8 π}{7}) = 4 sin (\frac{2 π}{7}) sin (\frac{4 π}{7}) sin (\frac{π}{7})$

Explanation:

$sin (\frac{2 π}{7}) + sin (\frac{4 π}{7}) + sin (\frac{8 π}{7})$

= $sin (\frac{8 π}{7}) + sin (\frac{2 π}{7}) + sin (\frac{4 π}{7})$

= $2 sin (\frac{8 π + 2 π}{2 \times 7}) cos (\frac{8 π - 2 π}{2 \times 7}) + sin (\frac{4 π}{7})$

= $2 sin (\frac{5 π}{7}) cos (\frac{3 π}{7}) + 2 sin (\frac{2 π}{7}) cos (\frac{2 π}{7})$

= $2 sin (π - \frac{5 π}{7}) cos (\frac{3 π}{7}) + 2 sin (\frac{2 π}{7}) cos (\frac{2 π}{7})$

= $2 sin (\frac{2 π}{7}) cos (\frac{3 π}{7}) + 2 sin (\frac{2 π}{7}) cos (π - \frac{5 π}{7})$

= $2 sin (\frac{2 π}{7}) [cos (\frac{3 π}{7}) - cos (\frac{5 π}{7})]$

= $2 sin (\frac{2 π}{7}) [2 sin (\frac{5 π + 3 π}{2 \times 7}) sin (\frac{5 π - 3 π}{2 \times 7})]$

= $4 sin (\frac{2 π}{7}) sin (\frac{4 π}{7}) sin (\frac{π}{7})$

Answer 2 · 2017-06-11T08:28:01

$sin (\frac{2 π}{7}) + sin (\frac{4 π}{7}) + sin (\frac{8 π}{7}) = \frac{\sqrt{7}}{2}$

Explanation:

Let:

$α = cos (\frac{2 π}{7}) + i sin (\frac{2 π}{7})$

Then:

$α^{7} = cos (2 π) + i sin (2 π) = 1$

So:

$0 = α^{7} - 1 = (α - 1) (α^{6} + α^{5} + α^{4} + α^{3} + α^{2} + α + 1)$

We find:

${(α + α^{2} + α^{4})}^{2}$

$= {(α)}^{2} + {(α^{2})}^{2} + {(α^{4})}^{2} + 2 (α α^{2} + α^{2} α^{4} + α^{4} α)$

$= α^{2} + α^{4} + α^{8} + 2 (α^{3} + α^{6} + α^{5})$

$= α + α^{2} + α^{4} + 2 (α + α^{2} + α^{3} + α^{4} + α^{5} + α^{6} - α - α^{2} - α^{4})$

$= α + α^{2} + α^{4} + 2 (- 1 - α - α^{2} - α^{4})$

$= - (α + α^{2} + α^{4}) - 2$

So $α + α^{2} + α^{4}$ is a root of:

$t^{2} + t + 2 = 0$

which we can solve by completing the square:

$0 = t^{2} + t + 2$

$0 = {(t + \frac{1}{2})}^{2} + \frac{7}{4}$

$0 = {(t + \frac{1}{2})}^{2} - {(\frac{\sqrt{7}}{2} i)}^{2}$

$0 = ((t + \frac{1}{2}) - \frac{\sqrt{7}}{2} i) ((t + \frac{1}{2}) + \frac{\sqrt{7}}{2} i)$

$0 = (t + \frac{1}{2} - \frac{\sqrt{7}}{2} i) (t + \frac{1}{2} + \frac{\sqrt{7}}{2} i)$

So:

$α + α^{2} + α^{4} = t = - \frac{1}{2} \pm \frac{\sqrt{7}}{2} i$

That is:

$cos (\frac{2 π}{7}) + cos (\frac{4 π}{7}) + cos (\frac{8 π}{7}) + i (sin (\frac{2 π}{7}) + sin (\frac{4 π}{7}) + sin (\frac{8 π}{7})) = - \frac{1}{2} \pm \frac{\sqrt{7}}{2} i$

Equating imaginary parts:

$sin (\frac{2 π}{7}) + sin (\frac{4 π}{7}) + sin (\frac{8 π}{7}) = \pm \frac{\sqrt{7}}{2}$

Which sign is correct?

Note that:

$sin (\frac{2 π}{7}) > 0$

$sin (\frac{4 π}{7}) > 0$

$sin (- \frac{2 π}{7}) < sin (\frac{- π}{7}) = sin (\frac{8 π}{7})$

Hence:

$sin (\frac{2 π}{7}) + sin (\frac{4 π}{7}) + sin (\frac{8 π}{7}) > 0$

So:

$sin (\frac{2 π}{7}) + sin (\frac{4 π}{7}) + sin (\frac{8 π}{7}) = \frac{\sqrt{7}}{2}$

How do you simplify sin(2π7)+sin(4π7)+sin(8π7) ?

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Explanation:

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How do you simplify $sin (\frac{2 π}{7}) + sin (\frac{4 π}{7}) + sin (\frac{8 π}{7})$ ?