Question

Fie a,b apartine lui R, astfel incat a+b=pi/3. Demonstrati ca sin2a-sin2b-sin(a-b)=0

Answer 1

$a,b\in \mathbb{R}\,\text{ a.i. }\, a+b = \dfrac{\pi}{3}$

$\\\sin 2a-\sin 2b -\sin(a-b) =\\ \\=2\sin \left(\dfrac{2a-2b}{2}\right)\cos\left(\dfrac{2a+2b}{2}\right) -\sin(a-b) \\ \\=2\sin(a-b)\cos(a+b) - \sin(a-b)\\ \\ =\sin(a-b)\left[2\cos(a+b)-1\right] \\ \\ =\sin(a-b)\left(2\cos\dfrac{\pi}{3}-1\right)\\ \\ =\sin(a-b)\left(2\cdot \dfrac{1}{2}-1\right)\\ \\ = \sin(a-b)(1-1)\\ \\ = \sin(a-b)\cdot 0\\ \\ = \boxed{0}$