Question 1

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

Question 2

Answer:

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

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

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

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