Question

demonstrați :
$\frac{1 + \cos(4x) }{ \sin(3x) - \sin(x) } = \frac{ \cos(x) }{ \sin(x) }$

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Answer 1

Răspuns:

$\frac{1+cos(4x)}{sin(3x)-sin(x)}=\frac{cos(2x)}{sin(x)}$

Fac separat 1+cos(4x) si sin(3x)-sin(x)

$1+cos(4x)=cos(0)+cos(4x)=2cos\frac{0+4x}{2}*cos\frac{0-4x}{2}=2cos(2x)*cos(-2x)$

$=2cos(2x)*cos(2x)=2cos^2(2x)$

$sin(3x)-sin(x)=2sin\frac{3x-x}{2}*cos\frac{3x+x}{2}=2sin(x)*cos(2x)$

Inlocuim:

$\frac{2cos^2(2x)}{2sin(x)*cos(2x)}=\frac{cos(2x)}{sin(x)}$ "Adevarat"