3

In my solution here, it was shown that $$\omega+\omega^2+\omega^4=-\frac 12\pm\frac{\sqrt7}2\qquad\qquad (\omega=e^{i2\pi/7})$$ from which we know that $$\sin \frac{2\pi}7+\sin \frac{4\pi}7-\sin \frac{6\pi}7=\Im (\omega+\omega^2+\omega^4)=\frac{\sqrt7}2$$ This can also be verified easily by computation.

By the same token it would appear that $$\cos \frac{2\pi}7+\cos \frac{4\pi}7-\cos \frac{6\pi}7=\Re (\omega+\omega^2+\omega^4)=\frac 12$$ However a quick computational check shows that the result is $1.3019...$ and not $\frac 12$.

Why is this so?

Community
  • 1
Hypergeometricx
  • 23,235

2 Answers2

2

In that answer there is the following equality $$\begin{align} \sin\frac{2\pi}7+\sin\frac{4\pi}7-\sin\frac{6\pi}7 &=\Im(\omega^1+\omega^2-\omega^3)\\ &=\Im(\omega^1+\omega^2+\omega^4)\\ &=\Im(S)\\ &=\dfrac{\sqrt{7}}{2}\\ \end{align}.$$ For the real part you have $$\begin{align} -\dfrac{1}{2}&= \Re(S)\\ &=\Re(\omega^1+\omega^2+\omega^4)\\ &=\Re(\omega^1+\omega^2+\omega^3)\\ &=\cos\frac{2\pi}7+\cos\frac{4\pi}7+\cos\frac{6\pi}7 \end{align}$$

Sfarla
  • 1,567
1

You're starting from the wrong assumption that $-\omega^3=\omega^4$, which is obviously wrong, because it would imply $\omega=-1$.

The true fact is that $$ -\sin\frac{6\pi}{7}=\sin\frac{8\pi}{7} $$ but $$ \cos\frac{6\pi}{7}=\cos\frac{8\pi}{7} $$ because $$ \frac{6\pi}{7}+\frac{8\pi}{7}=2\pi $$ and $$ \sin(2\pi-\alpha)=-\sin\alpha,\qquad \cos(2\pi-\alpha)=\cos\alpha $$

egreg
  • 244,946