Prove: $\sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}=\frac{\sqrt2}{2}$

So, you have

$$2\sin\frac{2\pi}{20}(\cos\frac{\pi}{20}+\sin\frac{\pi}{20})=2\sin\frac{2\pi}{20}\left(\frac{\cos^2\frac{\pi}{20}-\sin^2\frac{\pi}{20}}{\cos\frac{\pi}{20}-\sin\frac{\pi}{20}}\right)= \frac{\sqrt2}{2},$$ or $$ = \frac{2 \sin\frac{2\pi}{20}\cos\frac{2\pi}{20}}{\cos\frac{\pi}{20}-\sin\frac{\pi}{20}} = \frac{\sin\frac{\pi}{5}}{\cos\frac{\pi}{20}-\sin\frac{\pi}{20}}=\frac{\sqrt2}{2}.$$

As $\sin\frac{\pi}{20}= \sin\left(\frac{\pi}{4}-\frac{\pi}{5}\right) = \frac{\sqrt{2}}{2}(\cos\frac{\pi}{5}-\sin\frac{\pi}{5})$ and $\cos\frac{\pi}{20}= \frac{\sqrt{2}}{2}(\cos\frac{\pi}{5}+\sin\frac{\pi}{5})$, we have

$$ \frac{\sin\frac{\pi}{5}}{\frac{\sqrt{2}}{2}2 \sin\frac{\pi}{5}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}.$$

Hint: multiply both sides with $\frac{\sqrt 2}{2} $: $$\underbrace{\sin\frac{\pi}{20}+\cos\frac{\pi}{20}}+\underbrace{\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}}=\frac{\sqrt 2}{2}\;\;\;/\cdot \frac{\sqrt 2}{2} $$ so we have to prove: $$\sin\frac{3\pi}{10}-\sin\frac{\pi}{10}=\frac{1}{2} $$

Let $A,B,C,\ldots,T$ be points on a circle with diameter $1$ dividing it into $20$ equal arcs. Let $RG$ intersect $BO, BK, KF$ at $U,V,W$, respectively.

It is easy to get the following equalities $$\angle URO = \angle OUR = \angle BUV = \angle UVB = \angle WVK = \angle KWV = \angle FWG = \angle WGF = \frac 25 \pi,$$ so in particular $OR=OU$, $UB=VB$, $VK=WK$, and $WF=FG$.

Moreover \begin{align*} OR & = \sin \dfrac{3\pi}{20},\\ OB & = \sin \dfrac{7\pi}{20} = \cos \dfrac{3\pi}{20},\\ KB & = \sin \dfrac{9\pi}{20} = \cos \frac{\pi}{20}, \\ KF & = \sin \frac \pi 4 = \dfrac{\sqrt 2}{2}, \text{ and }\\ FG & = \sin \dfrac{\pi}{20}. \end{align*}

Therefore \begin{align*} \frac{\sqrt 2}{2} & = KF \\ & = KW + FG \\ & = KV + FG \\ & = BK - BV + FG \\ & = BK - BU + FG \\ & = BK - (OB - OU) + FG \\ & = BK - OB + OU + FG \\ & = BK - OB + OR + FG \\ & = FG + BK + OR - OB \\ & = \sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}. \end{align*}