Question

Să se arate că pentru x din domeniul de definiție are loc egalitatea:
$\frac{cosx + cos5x}{cos3x + cosx} - \frac{cos5x - cosx}{cos3x - cosx} = - 2$

Answer 1

$\displaystyle\\ \frac{\cos x + \cos 5x}{\cos 3x + \cos x} - \frac{\cos 5x - \cos x}{\cos 3x - \cos x} =\\\\ =\frac{\cos 5x + \cos x}{\cos 3x + \cos x} - \frac{\cos 5x - \cos x}{\cos 3x - \cos x} =\\\\ =\frac{2\cos\dfrac{5x+x}{2}\cos\dfrac{5x-x}{2}}{2\cos\dfrac{3x+x}{2}\cos\dfrac{3x-x}{2}} - \frac{-2\sin\dfrac{5x+x}{2}\sin\dfrac{5x-x}{2}}{-2\sin\dfrac{3x+x}{2}\sin\dfrac{3x-x}{2}} =$

$\displaystyle\\ =\frac{\cos 3x\cos 2x }{\cos 2x \cos x } - \frac{\sin 3x \sin 2x}{\sin 2x \sin x } =\\\\ =\frac{\cos 3x}{\cos x}-\frac{\sin 3x}{\sin x} =\\\\ =\frac{4\cos^3 x -3\cos x}{\cos x}-\frac{3\sin x-4\sin^3 x}{\sin x} =\\\\ =\frac{\cos x(4\cos^2 x -3)}{\cos x}-\frac{\sin x(3-4\sin^2 x)}{\sin x} =\\\\ =(4\cos^2 x -3) - (3-4\sin^2 x)=\\\\ =4\cos^2 x -3 - 3+4\sin^2 x=\\\\ =4(\sin^2 x + \cos^2 x)-6 = 4\cdot 1-6 = 4-6 = \boxed{\bf -2}$