Question

Prove that cosπ/5 + cos3π/5 = 1/2

Answer 1

Notice that equation $x^5-1=0$ has five roots, whose sum equals $0$ (from Vieta's theorem). It also means that their real parts also sum up to zero, i.e. $\cos\dfrac{2\pi}{5}+\cos\dfrac{4\pi}{5}-\cos\dfrac{\pi}{5}-\cos\dfrac{3\pi}{5}+1 = 0 \Leftrightarrow \cos\dfrac{\pi}{5}+\cos\dfrac{3\pi}{5} = 1+\cos\dfrac{2\pi}{5}+\cos\dfrac{4\pi}{5}$

By trig identities we have $\cos\dfrac{2\pi}{5}+\cos\dfrac{3\pi}{5}+\cos\dfrac{4\pi}{5} + \cos\dfrac{\pi}{5} = 2\cos\dfrac{\pi}{2}\cdot\ldots + 2\cos\dfrac{\pi}{2}\cdot\ldots = 0$ (the same result can be obtained by symmetry of the picture), so  $\cos\dfrac{\pi}{5}+\cos\dfrac{3\pi}{5} = 1+\cos\dfrac{2\pi}{5}+\cos\dfrac{4\pi}{5} = 1-\left(\cos\dfrac{\pi}{5}+\cos\dfrac{3\pi}{5} \right) \Leftrightarrow \cos\dfrac{\pi}{5}+\cos\dfrac{3\pi}{5} =1/2$.

Answer 2

Для доказательства этого равенства, мы можем использовать формулу для суммы косинусов двух углов:

cos(A+B) = cos(A)cos(B) - sin(A)sin(B)

Заменим A на π/5 и B на 3π/5:

cos(π/5 + 3π/5) = cos(π/5)cos(3π/5) - sin(π/5)sin(3π/5)

Упростим это выражение, зная, что cos(π/5) = cos(3π/5) и sin(π/5) = -sin(3π/5):

cos(4π/5) = cos^2(π/5) + sin^2(π/5)

Так как cos^2(π/5) + sin^2(π/5) = 1 (это следует из тригонометрического тождества), мы получаем:

cos(4π/5) = 1

Таким образом, мы доказали, что cos(π/5) + cos(3π/5) = 1/2.