Calculate $\cos(\frac{2\pi}{5})$ in terms of $\sin(\frac{\pi}{5})$

Question

I intend to calculate $\cos\left(\frac{2\pi}{5}\right)$ via the formula $$\sin\left(\frac{\pi}{5}\right)=\sqrt{\frac{1-\cos\left(\frac{2\pi}{5}\right)}{2}}$$ This could be expressed as:

$\sqrt{\frac{5-\sqrt{5}}{8}}=\sqrt{\frac{1-\cos\left(\frac{2\pi}{5}\right)}{2}}$. Squaring both sides and manipulate a bit, this will give $\cos\left(\frac{2\pi}{5}\right)=-\frac{1-\sqrt{5}}{4}$. However, the true result is $\frac{\sqrt{5}-1}{4}$. What is wrong with my derivation?

Answer 1

Note that

$$-\frac{1-\sqrt{5}}{4}=(-1)\cdot\frac{1-\sqrt{5}}{4}=\frac{(-1)\cdot1-(-1)\cdot(\sqrt{5})}{4}=\frac{\sqrt{5}-1}{4}$$

Answer 2

Relax, your answer is correct.

Just multiply the top by $-1$ and you will get $$\cos(\frac{2\pi}{5})=-\frac{1-\sqrt{5}}{4}= \frac{\sqrt{5}-1}{4}$$

Answer 3

You shouldn't have to square or manipulate whatever, if you used the relevant one of the duplication formulæ:

$$\cos 2\theta=\cos^2\theta-\sin^2\theta=2\cos^2\theta-1=1-2\sin^2\theta.$$