How do you simplify $\frac{cos 4 x - cos 2 x}{sin 4 x + sin 2 x}$ $(cos4x-cos2x)/(sin 4x + sin 2x)$ ?

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Answer 1 · 2018-05-19T03:58:07

$\Rightarrow - tan x$ $color(crimson)(=> - tan x$

$\frac{cos 4 x - cos 2 x}{sin 4 x + sin 2 x}$ $(cos 4x - cos 2x) / (sin 4x + sin 2x)$

$cos C - cos D = - 2 \frac{sin (C + D)}{2} \frac{sin (C - D)}{2}$ $cos C -cos D = -2 sin (C+D)/2 sin (C-D)/2$

$sin C + sin D = 2 \frac{sin (C + D)}{2} \frac{cos (C - D)}{2}$ $sin C +sin D = 2 sin (C +D)/2 cos (C -D)/2$

$N r = - 2 sin (\frac{4 x + 2 x}{2}) \cdot sin (\frac{4 x - 2 x}{2})$ $Nr = -2 sin ((4x+2x)/2) * sin ((4x-2x)/2)$

$N r = - 2 sin 3 x \cdot sin x$ $Nr = -2 sin 3x * sin x$

$D r = 2 sin (\frac{4 x + 2 x}{2}) \cdot cos (\frac{4 x - 2 x}{2})$ $Dr = 2 sin ((4x+2x)/2) * cos((4x-2x)/2)$

$D r = 2 sin 3 x \cdot cos x$ $Dr = 2 sin3x * cos x$

$(Nr) / (Dr) =- (cancel2 cancel(sin 3x) * sin x) / (cancel2 cancel(sin 3x) * cos x)$

$color(crimson)(=> - sin x / cos x = - tan x$

How do you simplify (cos4x-cos2x)/(sin 4x + sin 2x)cos4x−cos2xsin4x+sin2x(cos4x-cos2x)/(sin 4x + sin 2x)?