# de moivre ques involving tan

• August 4th 2011, 08:42 AM
blueyellow
de moivre ques involving tan
how does one go from

[Im(cos theta + i sin theta)^3]/[Re(cos theta + i sin theta)^3]

to

=[3cos^2 theta sin theta - sin^3 theta]/[cos^3 theta - 3 cos theta sin^2 theta]

I heard it's something to do with some triangle but I don't understand the details. Help would be much appreciated
• August 4th 2011, 08:55 AM
TheEmptySet
Re: de moivre ques involving tan
Quote:

Originally Posted by blueyellow
how does one go from

[Im(cos theta + i sin theta)^3]/[Re(cos theta + i sin theta)^3]

to

=[3cos^2 theta sin theta - sin^3 theta]/[cos^3 theta - 3 cos theta sin^2 theta]

I heard it's something to do with some triangle but I don't understand the details. Help would be much appreciated

You are probably thinking of Pascal's triangle. This is just a memory aid for the binomial theorem that states that

$(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}$

So in your case you would have in the numerator that

$(\cos (\theta) + i \sin (\theta))^3=\cos^3(\theta)+3i\cos^2(\theta)\sin( \theta )-3\cos(\theta)\sin^2(\theta)-i\sin^3(\theta)$

Now just take the imaginary part. You will need to do the same thing in the denominator.
• August 4th 2011, 09:00 AM
Plato
Re: de moivre ques involving tan
Quote:

Originally Posted by blueyellow
how does one go from
[Im(cos theta + i sin theta)^3]/[Re(cos theta + i sin theta)^3]
to
=[3cos^2 theta sin theta - sin^3 theta]/[cos^3 theta - 3 cos theta sin^2 theta]

I do not understand this question.
$\left[ {\text{Im}(\cos(\theta)+\mathbf{i}\sin(\theta))} \right]^3=\sin^3(\theta)$ and $\left[ {\text{Re}(\cos(\theta)+\mathbf{i}\sin(\theta))} \right]^3=\cos^3(\theta)$.

Thus your fraction just equals $\tan^3(\theta)$.
• August 4th 2011, 10:29 AM
HallsofIvy
Re: de moivre ques involving tan
I never thought I would say this but Plato is wrong. Perhaps he was thinking the cube is on the "cosine" and "sine" separately rather than cubing the entire complex number.

Since this is titled "DeMoivre", $cos(\theta)+ isin(\theta)= e^{i\theta}$, so that $[cos(\theta)+ i sin(\theta)]^3= e^{3i\theta}$. Going back to "rectangular form" that is $cos(3\theta)+ i sin(3\theta)$.

"Imaginary part over real part" is $\frac{sin(3\theta)}{cos(3\theta)}= tan(3\theta)$, not $tan^3(\theta)$.

But to answer the original question, $[cos(\theta)+ i sin(\theta)]^2= (cos^2(\theta)- sin^2(\theta))+ 2sin(\theta)cos(\theta)i$. Multiplying the real part of that by $cos(\theta)$ we get $cos^3(\theta)- sin^2(\theta)cos(\theta)$ and multiplying the imaginary part by $i sin(\theta)$ we get $-2sin^2(\theta)cos(\theta)$. That is, the real part of the cube is $cos^3(\theta)- 3sin^2(\theta)cos(\theta)$, the denominator of the fraction on the right.

Multiplying $cos^2(\theta)- sin^2(\theta)$ by $i sin(\theta)$ we get $(cos^2(\theta)sin(\theta)- sin^3(\theta))i$ and multiplying by $2sin(theta)cos(\theta) i$ by $cos(\theta)$ we get $2 sin(theta)cos^2(\theta)i$. That is, the imaginary part of the cube is $3cos^2(\theta)sin(\theta)- sin^3(\theta)$, the numerator of the fraction on the right.

You can also get that result from $tan(3\theta)= \frac{sin(3\theta)}{cos(3\theta)}$, using the identities $sin(a+ b)= sin(a)cos(b)+ cos(a)sin(b)$ and [tex]cos(a+ b)= cos(a)cos(b)- sin(a)sin(b). Taking $a= b= \theta$, $sin(2\theta)= sin(\theta)cos(\theta)+ cos(\theta)sin(\theta)= 2sin(\theta)cos(\theta)$ and $cos(2\theta)= cos^2(\theta)- sin^2(\theta)$. Then taking $a= 2\theta$, $b= \theta$, $sin(3\theta)= sin(\heta)cos(2\theta)+ cos(\theta)sin(2\theta)= sin(\theta)(cos^2(\theta)- sin^2(\theta))+ cos(\theta)(2sin(\theta)cos(\theta)= 3sin(\theta)cos^2(\theta)- sin^3(\theta)$, the numerator of the fraction on the right. $cos(3x)= cos(2x)cos(x)- sin(2x)sin(x)= (cos^2(\theta)- sin^2(\theta))cos(x)- (2sin(\theta)cos(\theta))sin(\theta)= cos^3(\theta)- 3cos^2(\theta)sin(\theta)$, the denominator of the fraction on the right.
• August 4th 2011, 11:01 AM
Plato
Re: de moivre ques involving tan
Quote:

Originally Posted by HallsofIvy
I never thought I would say this but Plato is wrong. Perhaps he was thinking the cube is on the "cosine" and "sine" separately rather than cubing the entire complex number.

Well the title says tangent.
Using a CAS $\text{Im}(z)^3$ means $[\text{Im}(z)]^3$ the cube of the imaginary part. If you notice I did add the extra grouping symbols in that reply.
So as I read it the imaginary part of the numerator is cubed and the real part of the denominator is cubed. That may be wrong but it is the convention I am use to. That does give us the tangent of the title.
• August 4th 2011, 11:23 AM
HallsofIvy
Re: de moivre ques involving tan
Thanks, that makes sense. I (who have never used a CAS) interpreted it differently.