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(cos(2x)-cos(4x))/(cos(2x)+cos(4x)) How does this fraction turn into (-2sin(3x)sin(-x))/(2cos(3x)cos(x))? Can you do a step by step proof to show how the 1st expression turns into the second expression?

Respuesta :

LammettHash
Recall the the angle sum identity for cosine.

[tex]\cos(\theta+\varphi)=\cos\theta\cos\varphi-\sin\theta\sin\varphi[/tex]
[tex]\cos(\theta-\varphi)=\cos\theta\cos\varphi+\sin\theta\sin\varphi[/tex]

Adding the two equations together gives

[tex]\cos(\theta+\varphi)+\cos(\theta-\varphi)=2\cos\theta\cos\varphi[/tex]

while subtracting gives

[tex]\cos(\theta+\varphi)-\cos(\theta-\varphi)=-2\sin\theta\sin\varphi[/tex]

Replacing [tex]\theta=3x[/tex] and [tex]\varphi=-x[/tex], you get

[tex]\dfrac{\cos2x-\cos4x}{\cos2x+\cos4x}=\dfrac{\cos(3x+(-x))-\cos(3x-(-x))}{\cos(3x+(-x))+\cos(3x-(-x))}=\dfrac{-2\sin3x\sin(-x)}{2\cos3x\cos(-x)}[/tex]

Finally, [tex]\cos(-x)=\cos x[/tex], so you get the second expression.

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