Question

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

Answer 1

1400 answer choice B

Answer 2

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

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

We get:

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

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

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

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

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

`=-\cot\left(x\right)`

Hence final answer is

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