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

ok, what I saw instantly is that:
$$\sin\frac{\pi}{20}+\sin\frac{3\pi}{20}=2\sin\frac{2\pi}{20}\cos\frac{\pi}{20}$$
and that,
$$\cos\frac{\pi}{20}-\cos\frac{3\pi}{20}=-2\sin\frac{2\pi}{20}\sin\frac{\pi}{20}$$
So, 
$$2\sin\frac{2\pi}{20}(\cos\frac{\pi}{20}-\sin\frac{\pi}{20})=\frac{\sqrt2}{2}=\sin\frac{5\pi}{20}$$
Unfortunately, I can't find a way to continue this, any ideas or different ways of proof?
*Taken out of the TAU entry exams (no solutions are offered)
 A: 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
$$ \frac{\sin\frac{\pi}{5}}{\frac{\sqrt{2}}{2}2 \sin\frac{\pi}{5}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}.$$
A: 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:
$$\sin\frac{3\pi}{10}-\sin\frac{\pi}{10}=\frac{1}{2} $$
A: 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*}
A: HINT: $$\left[\left(\sin\frac{\pi}{20}+\cos\frac{\pi}{20}\right)+\left(\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}\right)\right]^2$$ is equal to $$2+2\sin\frac{\pi}{20}\cos\frac{\pi}{20}-2\sin\frac{3\pi}{20}\cos\frac{3\pi}{20}+2\left(\sin\frac{\pi}{20}+\cos\frac{\pi}{20}\right)\left(\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}\right).$$ The double-angle and addition formulas for $\sin$ might come in handy...
A: Solution
Notice that 

$$\sin x \pm\cos x=\sqrt{2}\sin\left(x\pm\frac{\pi}{4}\right),~~~\forall x \in \mathbb{R}$$

Therefore,
\begin{align*}
&\sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}\\
=&\sqrt{2}\sin\left(\frac{\pi}{20}+\frac{\pi}{4}\right)+\sqrt{2}\sin\left(\frac{3\pi}{20}-\frac{\pi}{4}\right)\\
=&\sqrt{2}\left(\sin\frac{3\pi}{10}-\sin\frac{\pi}{10}\right)\\
=&2\sqrt{2}\sin \frac{\pi}{10}\cos\frac{\pi}{5}\\
=&2\sqrt{2}\cdot \frac{2\sin\dfrac{\pi}{10}\cos\dfrac{\pi}{10}\cos\dfrac{\pi}{5}}{2\cos\dfrac{\pi}{10}}\\
=&\sqrt{2}\cdot \frac{2\sin\dfrac{\pi}{5}\cos\dfrac{\pi}{5}}{2\cos\dfrac{\pi}{10}}\\
=&\sqrt{2}\cdot \frac{\sin\dfrac{2\pi}{5}}{2\cos\dfrac{\pi}{10}}\\
=&\sqrt{2}\cdot \frac{\cos\dfrac{\pi}{10}}{2\cos\dfrac{\pi}{10}}\\
=&\frac{\sqrt{2}}{2}.
\end{align*}

Note In fact, you may readily evaluate $\sin \dfrac{\pi}{10}$ and $\cos\dfrac{\pi}{5}$. As the figure shows, you may obtain $$\frac{AB}{BC}=\frac{BC}{CD},$$namely,$$\frac{x+2y}{x}=\frac{x}{2y},$$i.e. $$\left(\frac{x}{y}\right)^2-2\left(\frac{x}{y}\right)-4=0.$$Hence, $$\frac{x}{y}=1+\sqrt{5}.$$Therefore,$$\sin \dfrac{\pi}{10}=\frac{y}{x}=\frac{\sqrt{5}-1}{4},~~~\cos\dfrac{\pi}{5}=\frac{x+y}{x+2y}=\dfrac{\dfrac{x}{y}+1}{\dfrac{x}{y}+2}=\frac{\sqrt{5}+1}{4}.$$As a result,$$\sin \dfrac{\pi}{10}\cos\dfrac{\pi}{5}=\frac{1}{4}.$$

A: Since $\surd2\sin\frac14\pi=\surd2\cos\frac14\pi=1$, we may write the difference as$$\surd2\left(\cos\frac\pi4\cos\frac\pi{20}+\sin\frac\pi4\sin\frac\pi{20}\right)-\surd2\left(\cos\frac\pi4\cos\frac{3\pi}{20}-\sin\frac\pi4\sin\frac{3\pi}{20}\right)-\frac{\surd2}2$$$$=\frac{\surd2}2\left(2\cos\frac\pi5-2\cos\frac{2\pi}5-1\right)\qquad\qquad\qquad\qquad\qquad$$$$=\frac{\surd2}2\left(2\cos\frac\pi5\sin\frac\pi5-2\cos\frac{2\pi}5\sin\frac\pi5-\sin\frac\pi5\right)/\sin\frac\pi5$$$$=\frac{\surd2}2\left(\sin\frac{2\pi}5+\sin\frac\pi5-\sin\frac{3\pi}5-\sin\frac\pi5\right)/\sin\frac\pi5\qquad$$$$=0,$$since $\sin\frac35\pi=\sin(\pi-\frac35\pi)=\sin\frac25\pi$.
A: $$\sin9^\circ+\cos9^\circ+\sin27^\circ-\cos27^\circ$$
$$=\sin9^\circ+\sin(90-9)^\circ+\sin27^\circ+\sin(27-90)^\circ$$
Observe that $\sin5x=\sin45^\circ$  for $x=9^\circ,81^\circ,27^\circ,-63^\circ$
Now if $\sin5x=\sin45^\circ,$
$5x=180^\circ m+(-1)^m45^\circ$ where $m$ is any integer
$\implies x=36^\circ m+(-1)^m9^\circ$ where $m\equiv0,1,2,3,4\pmod5$
Again, $$\sin5x=5\sin x-20\sin^3x+16\sin^5x$$
$$\implies$$ the roots of $$16s^5-20s^3+5s-\sin45^\circ=0$$
are $x=36^\circ m+(-1)^m9^\circ$ where $m\equiv-1,0,1,2,3\pmod5$
$$\implies\sum_{m=-1}^3\sin(36^\circ m+(-1)^m9^\circ)=\dfrac0{16}$$
$$\implies\sum_{m=0}^3\sin(36^\circ m+(-1)^m9^\circ)=\sin(36^\circ\cdot-1+(-1)^{-1}9^\circ)=-\sin(-45^\circ)=?$$
