# Thread: [SOLVED] Complex Trig proof

1. ## [SOLVED] Complex Trig proof

*This is exactly how the question reads from the book

Using the fact that a complex equation is really two real equations, find the double angle formulas (for Sin2 $\theta\$, Cos2 $\theta\$ ) by using this equation: (e^i $\theta\$)^n = (Cos $\theta\$+ iSin $\theta\$)^n = Cos n $\theta\$ + iSin n $\theta\$

I know if you put in 2 for n and just exapand it out I can get it but I figured that there might be some other way to prove it. Just seems a bit simple but any help or insight would be greatly appriciated.

2. Hello,
Originally Posted by gidget
*This is exactly how the question reads from the book

Using the fact that a complex equation is really two real equations, find the double angle formulas (for Sin2 $\theta\$, Cos2 $\theta\$ ) by using this equation: (e^i $\theta\$)^n = (Cos $\theta\$+ iSin $\theta\$)^n = Cos n $\theta\$ + iSin n $\theta\$

I know if you put in 2 for n and just exapand it out I can get it but I figured that there might be some other way to prove it. Just seems a bit simple but any help or insight would be greatly appriciated.
Nope, you'll just have to expan $(\cos \theta + i \sin \theta)^2$

Note that $\cos(2 \theta)$ is the real part of it, and $\sin(2 \theta)$ the imaginary part.

3. Hello, gidget!

Using the fact that a complex equation is really two real equations,
find the double-angle formulas for $\sin2\theta,\;\cos2\theta$
by using this equation: $(e^{i\theta})^n \:=\: (\cos\theta + i\sin\theta)^n \:=\: \cos(n]\theta) + i\sin(n\theta)$

I know if you put in $n=2$ and just expand it out. I can get it.
But I figured that there might be some other way to prove it.
. .
I don't know of a simpler way.
I'm truly grateful for this method of cranking out multiple-angle identities.

If we want a formula for $\cos3\theta$, we'd normally proceed like this:

$\cos3\theta \;=\;\cos(\theta + 2\theta)$

. . . . $= \;\cos\theta\cos2\theta - \sin\theta\sin2\theta$

. . . . $= \;\cos\theta (2\cos^2\!\theta - 1) - \sin\theta(2\sin\theta\cos\theta)$

. . . . $= \;2\cos^3\!\theta - \cos\theta - 2\sin^2\!\theta\cos\theta$

. . . . $= \;2\cos^3\!\theta - \cos\theta - 2(1-\cos^2\!\theta)\cos\theta$

. . . . $= \;2\cos^3\!\theta - \cos\theta - 2\cos\theta + 2\cos^3\!\theta$

Therefore: . $\boxed{\cos3\theta \;=\;4\cos^3\!\theta - 3\cos\theta}$

Then we must repeat the process for $\sin3\theta.$

Using Demoivre's Formula, we have:

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

. . . . . . . . . . . . $=\;\cos^3\!\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\!\theta - i\sin^3\!\theta$

. . . . . . . . . . . . $= \;\left(\cos^3\!\theta - 3\cos\theta\sin^2\!\theta\right) + i\left(3\cos^2\!\theta\sin\theta - \sin^3\!\theta\right)$

. . . . . . . . . . . . $=\;\left(\cos^3\!\theta - 3\cos\theta[1-\cos^2\!\theta]\right) + i\left(3[1-\sin^2\!\theta]\sin\theta - \sin^3\!\theta\right)$

. . . . . . . . . . . . $= \;\left(4\cos^3\!\theta - 3\cos\theta\right) + i\left(3\sin\theta - 4\sin^3\!\theta\right)$

Equating real and imaginary components, we have:

. . . . $\begin{array}{ccc} \boxed{\cos3\theta \;=\; 4\cos^3\!\theta - 3\cos\theta} \\ \\[-3mm] \boxed{\sin3\theta \;=\; 3\sin\theta - 4\sin^3\theta} \end{array}$

And we have both formulas . . . with much less work!