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

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

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

So in your case you would have in the numerator that

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

Now just take the imaginary part. You will need to do the same thing in the denominator.

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.

$\displaystyle \left[ {\text{Im}(\cos(\theta)+\mathbf{i}\sin(\theta))} \right]^3=\sin^3(\theta) $ and $\displaystyle \left[ {\text{Re}(\cos(\theta)+\mathbf{i}\sin(\theta))} \right]^3=\cos^3(\theta) $.

Thus your fraction just equals $\displaystyle \tan^3(\theta)$.

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

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

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

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

You can also get that result from $\displaystyle tan(3\theta)= \frac{sin(3\theta)}{cos(3\theta)}$, using the identities $\displaystyle sin(a+ b)= sin(a)cos(b)+ cos(a)sin(b)$ and [tex]cos(a+ b)= cos(a)cos(b)- sin(a)sin(b). Taking $\displaystyle a= b= \theta$, $\displaystyle sin(2\theta)= sin(\theta)cos(\theta)+ cos(\theta)sin(\theta)= 2sin(\theta)cos(\theta)$ and $\displaystyle cos(2\theta)= cos^2(\theta)- sin^2(\theta)$. Then taking $\displaystyle a= 2\theta$, $\displaystyle b= \theta$, $\displaystyle sin(3\theta)= sin(\heta)cos(2\theta)+ cos(\theta)sin(2\theta)= sin(\theta)(cos^2(\theta)- sin^2(\theta))+ cos(\theta)(2sin(\theta)cos(\theta)= 3sin(\theta)cos^2(\theta)- sin^3(\theta)$, the numerator of the fraction on the right. $\displaystyle cos(3x)= cos(2x)cos(x)- sin(2x)sin(x)= (cos^2(\theta)- sin^2(\theta))cos(x)- (2sin(\theta)cos(\theta))sin(\theta)= cos^3(\theta)- 3cos^2(\theta)sin(\theta)$, the denominator of the fraction on the right.

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 $\displaystyle \text{Im}(z)^3$ means $\displaystyle [\text{Im}(z)]^3$ 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.

Re: de moivre ques involving tan

Thanks, that makes sense. I (who have never used a CAS) interpreted it differently.