How do you verify the identity: $- cot x = \frac{sin 3 x + sin x}{cos 3 x - cos x}$ $-cotx =(sin3x+sinx)/(cos3x-cosx)$ ?

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How do you verify the identity: $- cot x = \frac{sin 3 x + sin x}{cos 3 x - cos x}$ $-cotx =(sin3x+sinx)/(cos3x-cosx)$ ?

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Answer 1 · 2015-02-06T17:55:51

There are some formulas, named sum-to-product, that say:

$sin θ + sin ϕ = 2 sin (\frac{θ + ϕ}{2}) cos (\frac{θ - ϕ}{2})$ $sintheta+sinphi=2sin((theta+phi)/2)cos((theta-phi)/2)$ ,
$sin θ - sin ϕ = 2 cos (\frac{θ + ϕ}{2}) sin (\frac{θ - ϕ}{2})$ $sintheta-sinphi=2cos((theta+phi)/2)sin((theta-phi)/2)$ ,
$cos θ + sin ϕ = 2 cos (\frac{θ + ϕ}{2}) cos (\frac{θ - ϕ}{2})$ $costheta+sinphi=2cos((theta+phi)/2)cos((theta-phi)/2)$ ,
$cos θ + cos ϕ = - 2 sin (\frac{θ + ϕ}{2}) sin (\frac{θ - ϕ}{2})$ $costheta+cosphi=-2sin((theta+phi)/2)sin((theta-phi)/2)$ .

So:

$- cot x = \frac{2 sin (\frac{3 x + x}{2}) cos (\frac{3 x - x}{2})}{- 2 sin (\frac{3 x + x}{2}) sin (\frac{3 x - x}{2})}$ $-cotx=(2sin((3x+x)/2)cos((3x-x)/2))/(-2sin((3x+x)/2)sin((3x-x)/2))$ ;

$- cot x = - \frac{cos x}{sin x}$ $-cotx=-(cosx)/sinx$ .

That's all.

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How do you verify the identity: $- cot x = \frac{sin 3 x + sin x}{cos 3 x - cos x}$ $-cotx =(sin3x+sinx)/(cos3x-cosx)$ ?

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