Răspuns :
Răspuns:
a)
[tex]sin x+sin3x+sin5x=(sinx+sin5x)+sin3x =[/tex]
[tex]= 2sin\frac{x+5x}{2}*cos\frac{x-5x}{2} +sin3x<=>2sin\frac{6x}{2}*cos\frac{-4}{2}+sin3x[/tex]
[tex]=2sin3x*cos(-2x)+sin3x <=> sin3x(2cos2x+1)[/tex]
b)
[tex]1+cosx+cos2x=(1+cos2x)+cosx=(cos0+cos2x)+cosx[/tex]
[tex]=2cos\frac{0+2x}{2}*cos\frac{0-2x}{2}+cosx <=> 2cosx*cos(-x)+cosx[/tex]
[tex]=2cosx*cosx+cosx <=>2cosx^2+cosx <=>cosx(2cosx+1)[/tex]
[tex]\it =2sin\dfrac{x+5x}{2}cos\dfrac{x-5x}{2}+sin3x=2\sin3x\cos2x+\sin3x=\sin3x(2\cos2x+1)\\ \\ \rule{320}{0.8}\\ \\ \\ = 1+2cos^2x-1+cosx=2cos^2x+cosx=cosx(2cosx+1)[/tex]