Question

sa se calculeza suma a+b stiind ca sin(a+b+$\frac{ \pi }{6}$ )=-$\frac{1}{2}$ si sin(a+b-$\frac{ \pi }{6}$ )=$\frac{1}{2}$

Answer 1

Pentru rapiditate, notam a+b=x
Avem $sin(x+\dfrac{\pi}{6})=sinx\ cos \dfrac{\pi}{6}+cosx\ sin\dfrac{\pi}{6}\Rightarrow \dfrac{\sqrt3}{2}sinx+\dfrac12cosx=-\dfrac12$

Analog:

$sin(x-\dfrac{\pi}{6})=sinx\ cos \dfrac{\pi}{6}-cosx\ sin\dfrac{\pi}{6}\Rightarrow \dfrac{\sqrt3}{2}sinx-\dfrac12cosx=\dfrac12$

Adunam cele doua relatii si obtinem:
$\sqrt3\ sinx=0\Rightarrow sinx=0\Rightarrow x=a+b=k\pi;\ k\in\mathbb Z$