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Question

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

ok, lo que vi al instante es que:

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y eso,

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Entonces,$$\sin\frac{\pi}{20}+\sin\frac{3\pi}{20}=2\sin\frac{2\pi}{20}\cos\frac{\pi}{20}$ $

Desafortunadamente, no puedo encontrar una manera de continuar esto, ¿alguna idea o formas diferentes de prueba?

* Sacado de los exámenes de ingreso de TAU (no se ofrecen soluciones)

Answer 1

Así que tienes

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

Como$\sin\frac{\pi}{20}= \sin\left(\frac{\pi}{4}-\frac{\pi}{5}\right) = \frac{\sqrt{2}}{2}(\cos\frac{\pi}{5}-\sin\frac{\pi}{5})$ y$\cos\frac{\pi}{20}= \frac{\sqrt{2}}{2}(\cos\frac{\pi}{5}+\sin\frac{\pi}{5})$, tenemos

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Answer 2

Sugerencia: multiplique ambos lados con$\frac{\sqrt 2}{2} $:$$\underbrace{\sin\frac{\pi}{20}+\cos\frac{\pi}{20}}+\underbrace{\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}}=\frac{\sqrt 2}{2}\;\;\;/\cdot \frac{\sqrt 2}{2} $ $, por lo que debemos probar:$$\sin\frac{3\pi}{10}-\sin\frac{\pi}{10}=\frac{1}{2} $ $

Answer 3

Deje$A,B,C,\ldots,T$ puntos en un círculo con el diámetro$1$ dividiéndolo en$20$ arcos iguales. Permita que$RG$ se interseque$BO, BK, KF$ en$U,V,W$, respectivamente.

Es fácil obtener las siguientes igualdades$$\angle URO = \angle OUR = \angle BUV = \angle UVB = \angle WVK = \angle KWV = \angle FWG = \angle WGF = \frac 25 \pi,$ $, en particular$OR=OU$,$UB=VB$,$VK=WK$ y$WF=FG$.

Además, \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 *}

Por lo tanto, \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 *}

Answer 4

SUGERENCIA: $$\left[\left(\sin\frac{\pi}{20}+\cos\frac{\pi}{20}\right)+\left(\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}\right)\right]^2$$ 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 $ puede ser útil ...

Answer 5

Solución

Darse cuenta de

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Por lo tanto, \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 *}

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Nota De hecho, puede evaluar fácilmente$$\sin x \pm\cos x=\sqrt{2}\sin\left(x\pm\frac{\pi}{4}\right),~~~\forall x \in \mathbb{R}$ y$\sin \dfrac{\pi}{10}$. Como muestra la figura, puede obtener$\cos\dfrac{\pi}{5}$$$\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}.$% # ps