Calculate expression: $\cos\alpha-\cos2\alpha$ I attempt calculate this expression:
$$\cos\dfrac{\pi}{5}-\cos\dfrac{2\pi}{5}$$
Please, help me, somebody. Thanks.
 A: $\cos\dfrac{\pi}{5}=\cos36^\circ$ and the next is $\cos\dfrac{2\pi}{5}=\cos72^\circ$. Now so by converting into $\sin$ we have to do
$$\sin54^\circ-\sin18^\circ,$$
which can be done by following. 
Let $\alpha=18^\circ$, therefore $5\alpha=90^\circ$. So $\sin2\alpha=\sin(90^\circ-3\alpha)$, so $\sin2\alpha=\cos3\alpha$. Expanding both sides, and converting to $\sin$ we have:
$$4\sin^{2}\alpha+3\sin\alpha-1=0.$$ Hence, $\sin\alpha=\dfrac{\sqrt{5}-1}{4}$. Similarly find 
$$\sin3\alpha=\dfrac{\sqrt{5}+1}{4}.$$
So answer is 
$$\dfrac{\sqrt{5}+1}{4}-\dfrac{\sqrt{5}-1}{4}=\dfrac{1}{2}.$$
A: Let $\alpha=\pi/5$ for simplicity. By De Moivre formulas,
$$
(\cos\alpha+i\sin\alpha)^5=\cos\pi+i\sin\pi=-1
$$
and, if we set $z=\cos\alpha+i\sin\alpha$, $z^5+1=0$ that means
$$
(z+1)(z^4-z^3+z^2-z+1)=0
$$
Since $z\ne-1$, the second factor vanishes. Dividing by $z^2$ we get
$$
z^2+\frac{1}{z^2}-z-\frac{1}{z}+1=0
$$
And, recalling that $(z+\frac{1}{z})^2=z^2+2+\frac{1}{z^2}$,
$$
\left(z+\frac{1}{z}\right)^{\!2}-\left(z+\frac{1}{z}\right)-1=0
$$
Now
$$
z+\frac{1}{z}=z+\bar{z}=2\cos\alpha
$$
so $2\cos\alpha$ is the positive root of the equation $t^2-t-1=0$ and thus
$$
2\cos\alpha=\frac{1+\sqrt{5}}{2}
$$
Now compute
\begin{align}
\cos\alpha-\cos2\alpha
&=\cos\alpha-2\cos^2\alpha+1\\[6px]
&=\frac{1+\sqrt{5}}{4}-2\frac{1+2\sqrt{5}+5}{16}+1\\[6px]
&=\frac{1+\sqrt{5}-3-\sqrt{5}+4}{4}\\[6px]
&=\frac{1}{2}
\end{align}
A: One can find $\cos(\pi/5)$ in radicals as described here:     
http://mathworld.wolfram.com/TrigonometryAnglesPi5.html
Once we know $\cos(\pi/5)$, we can just use the formula
$\cos(2\alpha) = 2\cos^2(\alpha)-1$
to find the value of $\cos(2\pi/5)$.    
