# Thread: Trig Proof/Verification : cot 2x = (1+cos4x)/sin4x

1. ## Trig Proof/Verification : cot 2x = (1+cos4x)/sin4x

Sorry, another one.

If you could just explain how to split up the cos4x and sin4x, that would be helpful as well!

Thank you:

cot 2x = (1+cos4x)/sin4x

2. Originally Posted by SMA777
Sorry, another one.

If you could just explain how to split up the cos4x and sin4x, that would be helpful as well!

Thank you:

cot 2x = (1+cos4x)/sin4x
Use the identities $\displaystyle \cos{2x} = 2\cos^2{x} - 1$ and $\displaystyle \sin{2x} = 2\sin{x}\cos{x}$.

$\displaystyle \frac{1 + \cos{4x}}{\sin{4x}} = \frac{1 + \cos{2\times 2x}}{\sin{2\times 2x}}$

$\displaystyle = \frac{1 + 2\cos^2{2x} - 1}{2\sin{2x}\cos{2x}}$

$\displaystyle = \frac{2\cos^2{2x}}{2\sin{2x}\cos{2x}}$

$\displaystyle = \frac{\cos{2x}}{\sin{2x}}$

$\displaystyle = \cot{2x}$.

3. In your very first step, is it cos(2) * 2x or cos(2*2x), because the latter makes more sense, but in that case the step after that does not ...

Oh wait! I got it - becuase it's 2x the x in the Double Angle Formula is TWO x as well. Ah. Thank you!

4. Originally Posted by SMA777
In your very first step, is it cos(2) * 2x or cos(2*2x), because the latter makes more sense, but in that case the step after that does not ...

Oh wait! I got it - becuase it's 2x the x in the Double Angle Formula is TWO x as well. Ah. Thank you!
Yes it's $\displaystyle \cos{(2\times 2x)}.$

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# 1 cos4x formula

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