# Proof Tan 3x is = Sin 3x/ Cos 3x using De Moivre's theorem

##### New member
i have been able to do the following
But I'm stuck in simplifying the equation
See my working below
$$\displaystyle \cos3x+i\sin3x=\cos^3x+3i\cos^2x\sin x-3\cos x\sin^2x-i\sin^3x$$

\displaystyle \begin{align} \sin3x&=3\cos^2x\sin x-\sin^3x \\[6px] \cos3x&=\cos^3x-3\cos x\sin^2x \\[12px] \tan3x&=\frac{3\cos^2x\sin x-\sin^3x}{\cos^3x-3\cos x\sin^2x} \end{align}

$$\displaystyle \tan(3x)=\frac{3\tan(x)-\tan^3(x)}{1-3\tan^2(x)}$$

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#### Jomo

##### Elite Member
Look at your last two equations and try to match up the right hand sides. For example the 1in the last equation should (hopefully!!) match up with cos3x above. So what do you do to cos3x to get 1? Do the same to the entire numerator and the entire denominator!

#### Dr.Peterson

##### Elite Member
It's not clear what the goal is. You seem to have used what you said you want to prove! And that is essentially a definition anyway.

Or maybe I'm misinterpreting your work, as you didn't say in words what you are doing at each step.

Please state the wording of the assignment.

##### New member
Look at your last two equations and try to match up the right hand sides. For example the 1in the last equation should (hopefully!!) match up with cos3x above. So what do you do to cos3x to get 1? Do the same to the entire numerator and the entire denominator!
You divide by $$\displaystyle cos 3x$$

I'm guessing the last equation is the "actual" proof

I was expecting a $$\displaystyle \frac {Sin 3x }{ Cos 3x}$$
Which would be like proving De Moivre's theorem.

#### Jomo

##### Elite Member
You divide by $$\displaystyle cos 3x$$

I'm guessing the last equation is the "actual" proof

I was expecting a $$\displaystyle \frac {Sin 3x }{ Cos 3x}$$
Which would be like proving De Moivre's theorem.
No!, cos3x/cox 3x is NOT 1. You know this! cos3x/cos3x = 1!!

Just divide by cos3x

My mistake I meant to write $$\displaystyle cos ^3 (x)$$