Question

Simplify: cos2x-cos4 all over sin2x + sin 4x

Answer 1

`(\cos\left(2x\right)-\cos\left(4x\right))/(\sin\left(2x\right)+\sin\left(4x\right))=\tan\left(x\right)`

Explanation:

`(\cos\left(2x\right)-\cos\left(4x\right))/(\sin\left(2x\right)+\sin\left(4x\right))`

Apply formula:

`\cos\left(A\right)-\cos\left(B\right)=-2\cdot\sin\left((A+B)/(2)\right)\cdot\sin\left((A-B)/(2)\right)` and

`\sin\left(A\right)+\sin\left(B\right)=2\cdot\sin\left((A+B)/(2)\right)\cdot\sin\left((A-B)/(2)\right)`

We get:

`=(-2\cdot\sin\left((2x+4x)/(2)\right)\cdot\sin\left((2x-4x)/(2)\right))/(2\cdot\sin\left((2x+4x)/(2)\right)\cdot\cos\left((2x-4x)/(2)\right))`

`=(-\sin\left((2x-4x)/(2)\right))/(\cos\left((2x-4x)/(2)\right))`

`=(-\sin\left((-2x)/(2)\right))/(\cos\left((-2x)/(2)\right))`

`=(-\sin\left(-x\right))/(\cos\left(-x\right))`

`=(-\cdot-\sin\left(x\right))/(\cos\left(x\right))`

`=(\sin\left(x\right))/(\cos\left(x\right))`

`=\tan\left(x\right)`

Hence final answer is

`(\cos\left(2x\right)-\cos\left(4x\right))/(\sin\left(2x\right)+\sin\left(4x\right))=\tan\left(x\right)`