Răspuns:
Explicație pas cu pas:
[tex]\frac{sinx+2sin2x+sin3x}{cos3x-cosx}=\frac{(sin3x+sinx)+2sin2x}{cos3x-cosx}=\frac{2sin\frac{3x+x}{2}cos\frac{3x-x}{2}+2sin2x}{-2sin\frac{3x+x}{2}sin\frac{3x-x}{2}}=\\=\frac{2sin2xcosx+2sin2x}{-2sin2xsinx}=\frac{2sin2x(cosx+1)}{-2sin2xsinx}=\frac{cosx+1}{-sinx}=\frac{(1+cosx)}{-2sin\frac{x}{2}cos\frac{x}{2}}=\\=\frac{2cos^{2}\frac{x}{2} }{-2sin\frac{x}{2}cos\frac{x}{2}}=-\frac{cos\frac{x}{2}}{sin\frac{x}{2}}=-ctg\frac{x}{2}\\[/tex]
