Question

(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?

Answer 1

Recall the the angle sum identity for cosine.


`\cos(\theta+\varphi)=\cos\theta\cos\varphi-\sin\theta\sin\varphi`

`\cos(\theta-\varphi)=\cos\theta\cos\varphi+\sin\theta\sin\varphi`

Adding the two equations together gives


`\cos(\theta+\varphi)+\cos(\theta-\varphi)=2\cos\theta\cos\varphi`

while subtracting gives


`\cos(\theta+\varphi)-\cos(\theta-\varphi)=-2\sin\theta\sin\varphi`

Replacing
`\theta=3x` and
`\varphi=-x`, you get


`(\cos2x-\cos4x)/(\cos2x+\cos4x)=(\cos(3x+(-x))-\cos(3x-(-x)))/(\cos(3x+(-x))+\cos(3x-(-x)))=(-2\sin3x\sin(-x))/(2\cos3x\cos(-x))`

Finally,
`\cos(-x)=\cos x`, so you get the second expression.