Proving identities using demoivre's formula

[Drain Brain]

prove the following identities
$\cos(2\theta)=\cos^{2}(\theta)-\sin^{2}(\theta)$

$\sin(2\theta)=2\cos(\theta)\sin(\theta)$our Demoivre's formula says $z=r\left(\cos(\theta)+i\sin(\theta)\right)$

I don't know how use it to prove the identities above please help me get started.

regards!

 

[Prove It]

prove the following identities
$\cos(2\theta)=\cos^{2}(\theta)-\sin^{2}(\theta)$

$\sin(2\theta)=2\cos(\theta)\sin(\theta)$our Demoivre's formula says $z=r\left(\cos(\theta)+i\sin(\theta)\right)$

I don't know how use it to prove the identities above please help me get started.

regards!

The reason you don't know what to do is because you haven't even got DeMoivre's Formula written down correctly. It's actually:

If $\displaystyle \begin{align*} z = r \left[ \cos{ \left( \theta \right) } + \mathrm{ i }\sin{ \left( \theta \right) } \right] \end{align*}$ then $\displaystyle \begin{align*} z^n = r^n \left[ \cos{ \left( n\,\theta \right) } + \mathrm{ i }\sin{ \left( n\,\theta \right) } \right] \end{align*}$

So can you think of two different ways to evaluate $\displaystyle \begin{align*} z^2 \end{align*}$? (One is obviously using DeMoivre's Theorem...)

 

[Deveno]

I Just wanted to add here some general observations about complex numbers (they are fascinating!).

First, complex numbers have a "dual identity":

Algebraically, they are FORMAL sums of real numbers and imaginary numbers:

$z = x + yi$ (some books use $j$ instead of $i$, that's ok).

Geometrically, they are "points in the Argand plane":

$z = (x,y)$.

So when are two complex numbers $z,w$ equal? If $z = w$, and:

$z = x +yi,\ w = x' + y'i$, then $x = x'$ and $y = y'$, that is: their real and imaginary "parts" are equal.

This is the SAME criterion we have for two points $(x,y) = (x',y')$ being equal in the plane.

You should keep this uppermost in your mind, because two complex numbers might BE the same, and not LOOK the same.

Algebraically, we add complex numbers like so:

$(a + bi) + (c + di) = (a + c) + (bi + di) = (a + c) + (b + d)i$.

Geometrically, we add complex numbers like vectors:

$(a,b) + (c,d) = (a+c,b+d)$ (this makes a nice parallelogram, if you draw it).

Algebraically, we multiply two complex numbers like so:

$(a + bi)(c + di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc) i$ (since $i^2 = -1$).

Geometrically, we "change to polar coordinates":

$(x,y) = (r \cos\theta, r\sin\theta)$ and then:

$(x,y)\ast(x',y') = (r\cos\theta,r\sin\theta)\ast(r'\cos\theta',r'\sin\theta')$

$ = (rr'\cos(\theta+\theta'),rr'\sin(\theta+\theta'))$

that is, we multiply the magnitudes (dilation), and add the angles (rotation).

For this reason, complex numbers are often called "rotation-dilations".

DeMoivre's theorem, then, is just the "repeated application" of this when $(x,y) = (x',y')$, in other words a special case of a more general theorem:

The algebraic multiplication, and the geometric multiplication are the same.

The way to see this is true, most simply, is to use a THIRD view of the complex numbers as a "referee".

Since complex numbers can be viewed as "points" in the plane, and multiplication by one complex number to another, yields a third complex number, we have a function:

$f_z: \Bbb C \to \Bbb C$ for every complex number $z$:

$f_z(w) = zw$.

The distributive law for complex numbers tells us this is a LINEAR function, which can thus be represented by a 2x2 matrix.

Which matrix might it be? Suppose $z = a+bi (= (a,b))$, and $w = 1 = 1 + 0i$. Then $zw = z$. So if our matrix is $M_z$, we know that:

$M_z(1,0) = (a,b)$.

So the first column of $M_z$ is:

$\begin{bmatrix}a\\b \end{bmatrix}$

Now if $w = i$, then $zi = (a+bi)i = ai + bi^2 = -b + ai$. Remember $i = 0 + 1i = (0,1)$, so the SECOND column of $M$ is:

$\begin{bmatrix}-b\\a \end{bmatrix}$.

So the matrix corresponding to $z = a+bi$, is:

$M_z = \begin{bmatrix}a&-b\\b&a \end{bmatrix}$ <--our 3rd view of complex numbers, as 2x2 matrices!

We can verify that this gives $zw$ when applied to $w = c + di$:

$M_z(c,d) = \begin{bmatrix}a&-b\\b&a \end{bmatrix}\begin{bmatrix}c\\d \end{bmatrix}$

$= \begin{bmatrix}ac - bd\\ad+bc \end{bmatrix} = (ac-bd,ad+bc) = (ac-bd) + (ad+bc)i$, as before.

The "cool part" is what happens when we write $z$ in polar coordinates: $z = r\cos\theta + i\sin\theta$, where we get the matrix:

$M_z = \begin{bmatrix}r\cos\theta&-r\sin\theta\\r\sin\theta&r\cos\theta \end{bmatrix}$

$= r\begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{bmatrix}$

$= \begin{bmatrix}r&0\\0&r \end{bmatrix}\begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{bmatrix}$

where the second matrix is clearly "rotation by $\theta$" and the first is "stretching by $r$".

Interestingly enough, this explains why "a negative times a negative is a positive". Seen as a complex number, what -1 does is "rotate by 180 degrees", and if you do this twice, you rotate by 360 = 0 degrees, leaving you pointed back in the positive direction (the positive $x$-axis).

It also explains why $\sqrt{-1}$ isn't on the $x$-axis (why it isn't real), "halfway to 180 degrees" is only 90 degrees, and numbers on the $y$-axis are "off the line" (of the $x$-axis).

When seen this way, it seems that the "problem" with ordinary real numbers, is that there are actually "more directions" than just "left/right", complex numbers free magnitudes from the "tyranny of the line".

 

[soroban]

Hello, Drain Brain!

Using DeMoivre's formula, prove the following identities:

. . $\cos(2\theta)\:=\:\cos^{2}(\theta)-\sin^{2}(\theta)$

. . $\sin(2\theta)\:=\:2\cos(\theta)\sin(\theta)$

DeMoivre's formula: $\: (\cos\theta + i\sin\theta)^n \;=\;\cos(n\theta) + i\sin(n\theta)$

Let $n=2\!:$

$\quad \cos(2\theta ) +i\sin(2\theta) \;=\;(\cos\theta + i\sin\theta)^2$

$\quad\cos(2\theta) + i\sin(2\theta) \;=\; \cos^2\!\theta + 2i\sin\theta\cos\theta - \sin^2\!\theta$

$\quad\cos(2\theta) + i\sin(2\theta) \;=\;(\cos^2\!\theta- \sin^2\!\theta) + (2\sin\theta\cos\theta)i $Equate real and imaginary components:

$\quad \begin{Bmatrix}\cos(2\theta) &=& \cos^2\!\theta - \sin^2\!\theta \\
\sin(2\theta) &=& 2\sin\theta\cos\theta \end{Bmatrix}$

 

[soroban]

Here is a procedure I "discovered" while in college.Suppose we want the formula for $\tan(5x)$ in terms of $\tan(x).$Consider the terms of $(1 + \tan x)^5$

$\qquad 1,\;5\tan x,\;10\tan^2\!x,\;10\tan^3\!x,\;5\tan^4\!x,\; \tan^5\!x$Write the terms alternately in the numerator and denominator:

$\qquad \dfrac{1 \quad 10\tan^2\!x \quad 5\tan^4\!x}{5\tan x \quad 10\tan^3\!x \quad \tan^5\!x} $Insert alternating signs in the numerator and in the denominator:

$\qquad \dfrac{1 \,{\color{red}-}\,10\tan^2\!x\,{\color{red}+}\,5\tan^4\!x}{5\tan x\,{\color{red}-}\,10\tan^3\!x\,{\color{red}+} \tan^5\!x}$Therefore: $\:\tan(5x) \;=\; \dfrac{1 - 10\tan^2\!x + 5\tan^4\!x}{5\tan x - 10\tan^3\!x + \tan^5\!x}$

[Yes, this procedure is based on DeMoivre's formula.]