qwen2-math-1.5b-instruct

 >>> prove that cos(2*Pi/5)+cos(4*Pi/5)=-1/2

To prove that $\cos \left(\frac{2\pi }{5}\right)+\cos \left(\frac{4\pi }{5}\right)=-\frac{1}{2}$, we can use

properties of roots of unity and the symmetry of the cosine function.

First, consider the fifth roots of unity. These are the solutions to the equation $z^5=1$. The fifth roots of

unity are given by:

$$\begin{array}{}\text{(1)}& 1,\omega ,{\omega }^2,{\omega }^3,{\omega }^4,\end{array}$$

where $\omega =e^{2\pi i/5}$.

The sum of all the fifth roots of unity is zero:

$$\begin{array}{}\text{(2)}& 1+\omega +{\omega }^2+{\omega }^3+{\omega }^4=0.\end{array}$$

We can rewrite this equation as:

$$\begin{array}{}\text{(3)}& \omega +{\omega }^2+{\omega }^3+{\omega }^4=-1.\end{array}$$

Next, we use the fact that $\omega =e^{2\pi i/5}$ and its complex conjugate is  \omega^4\). The real part of $\omega +{\omega }^4$ is $2\cos \left(\frac{2\pi }{5}\right)$, and the real

part of ${\omega }^2+{\omega }^3$ is $2\cos \left(\frac{4\pi }{5}\right)$.

Thus, we can separate the sum of the roots into their real parts:

$$\begin{array}{}\text{(4)}& 2\cos \left(\frac{2\pi }{5}\right)+2\cos \left(\frac{4\pi }{5}\right)=-1.\end{array}$$

Dividing both sides by 2, we get:

$$\begin{array}{}\text{(5)}& \cos \left(\frac{2\pi }{5}\right)+\cos \left(\frac{4\pi }{5}\right)=-\frac{1}{2}.\end{array}$$

Therefore, the proof is complete and the answer is:

$$\begin{array}{}\text{(6)}& -\frac{1}{2}\end{array}$$