Question

Arătaţi că sin$\frac{5pi}{12}$ + cos $\frac{11pi}{12}$ = 0.

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

$\it \dfrac{5\pi}{12} <\dfrac{6\pi}{12} \Rightarrow \dfrac{5\pi}{12}< \dfrac{\pi}{2} \Rightarrow \dfrac{5\pi}{12}\in\Big(0,\ \dfrac{\pi}{2} \Big)\Rightarrow sin \dfrac{5\pi}{12}=cos\Big( \dfrac{\pi}{2}- \dfrac{5\pi}{12} \Big) =cos \dfrac{\pi}{12} \\ \\ \\ \left.\begin{aligned}\dfrac{11\pi}{12} < \dfrac{12\pi}{12}\Rightarrow \dfrac{11\pi}{12}<\pi\\ \\ Evident,\ \dfrac{11\pi}{12}> \dfrac{\pi}{2}\end{aligned}\right\} \Rightarrow cos \dfrac{11\pi}{12}=-cos\Big(\pi- \dfrac{11\pi}{12}\Big )\Rightarrow$

$\it \Rightarrow cos\dfrac{11\pi}{12}=-cos\dfrac{\pi}{12}$

Deci, relația din enunț este echivalentă cu:

$\it cos\dfrac{\pi}{12}-cos\dfrac{\pi}{12}=0\ \ (Adev\breve{a}rat\breve{a})$