4 Sine Pi over 10 by Cosine Pi over 5/Proof 2

Proof
First, we must rewrite the so that it reaches a form that is able to be substituted for what we determine is true through identities.

Now that we have reduced the expression to just one trigonometric function of a single angle, we can use the fact that a multiple of the angle is $\dfrac \pi 2$, whose cosine is $0$, in order to solve for $\cos \dfrac \pi {10}$.

Having realized that the above process is the same as the derivation of the expansion formula for $\cos {5 \theta}$, we can generalize this expression and then plug in $\dfrac \pi {10}$.

We were able to divide off the $u$ from both sides of that last line because we know that $\cos \dfrac \pi {10}$ cannot possibly equal $0$. From here, we use the quadratic formula to solve for $u^2$.

Taking the square root of both sides, we can solve for $u$.

We know that $\dfrac \pi {10}$ is in the first quadrant, so we can change the exterior $±$ sign to a $+$ sign:

We know that $\dfrac \pi {10}$ is a lot closer to $0$ than to $\dfrac \pi 2$:

$\dfrac \pi {10} - 0 < \dfrac \pi 2 - \dfrac \pi {10}$

Therefore, the larger of the possible interior quantities of the radical should be correct.

Now that we have evaluated $\cos \dfrac \pi {10}$, we can substitute it for the most simplified expression from above with only $\cos \dfrac \pi {10}$):

Note

 * In that line, we used a double-angle formula for cosine, a double-angle formula for sine, and the triple-angle formulas for each, which can be derived very easily with the angle-sum formulas for either using $2 \theta$ and $\theta$. $\cos {3 \theta} = 4 \cos^3 \theta - 3 \cos \theta$ and $\sin {3 \theta} = 3 \sin \theta - 4 \sin^3 \theta$.