# 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)

• $\cos(\frac{\pi}{20})-\cos(\frac{3\pi}{20})=2\sin(\frac{2\pi}{20})\sin(\frac{\pi}{20})$; not $-2\sin(\frac{2\pi}{20})\sin(\frac{\pi}{20})$. – Roronoa Zoro Jul 21 '18 at 13:45

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

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

• Nicely done. +1. – prog_SAHIL Jul 21 '18 at 14:17

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*}

• (+1) Nicely done! :) – TheSimpliFire Jul 22 '18 at 15:16

# 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,$$\sin \dfrac{\pi}{10}\cos\dfrac{\pi}{5}=\frac{1}{4}.$$ 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...

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 $\sin\frac35\pi=\sin(\pi-\frac35\pi)=\sin\frac25\pi$.

$$\sin9^\circ+\cos9^\circ+\sin27^\circ-\cos27^\circ$$

$$=\sin9^\circ+\sin(90-9)^\circ+\sin27^\circ+\sin(27-90)^\circ$$

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 $m\equiv-1,0,1,2,3\pmod5$

$$\implies\sum_{m=-1}^3\sin(36^\circ m+(-1)^m9^\circ)=\dfrac0{16}$$

$$\implies\sum_{m=0}^3\sin(36^\circ m+(-1)^m9^\circ)=\sin(36^\circ\cdot-1+(-1)^{-1}9^\circ)=-\sin(-45^\circ)=?$$