Question

Completați pentru a obtine un produs:

a) sin x + sin 3x + sin 5x = ( sin x + sin 5x ) + sin 3 x = ...

b)1 + cos x + cos 2x = ( 1 + cos 2 x ) + cos x = ...
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Answer 1

Răspuns:

a)

$sin x+sin3x+sin5x=(sinx+sin5x)+sin3x =$

$= 2sin\frac{x+5x}{2}*cos\frac{x-5x}{2} +sin3x<=>2sin\frac{6x}{2}*cos\frac{-4}{2}+sin3x$

$=2sin3x*cos(-2x)+sin3x <=> sin3x(2cos2x+1)$

b)

$1+cosx+cos2x=(1+cos2x)+cosx=(cos0+cos2x)+cosx$

$=2cos\frac{0+2x}{2}*cos\frac{0-2x}{2}+cosx <=> 2cosx*cos(-x)+cosx$

$=2cosx*cosx+cosx <=>2cosx^2+cosx <=>cosx(2cosx+1)$

Answer 2

$\it =2sin\dfrac{x+5x}{2}cos\dfrac{x-5x}{2}+sin3x=2\sin3x\cos2x+\sin3x=\sin3x(2\cos2x+1)\\ \\ \rule{320}{0.8}\\ \\ \\ = 1+2cos^2x-1+cosx=2cos^2x+cosx=cosx(2cosx+1)$