Please help me solve this problem.
if a=cos2alpha + isin2alpha
b=cos2beta + isin2beta
prove that,
a-b / a+b = itan (alpha-beta)
Assuming "2alpha" is $\displaystyle 2 \alpha$, then
$\displaystyle a = e^{2i \alpha}$
$\displaystyle b = e^{2i \beta}$
Then, $\displaystyle \frac{a-b}{a+b} = \frac{e^{2i \alpha} - e^{2i \beta}}{e^{2i \alpha} + e^{2i \beta}}$. Can you take it from there?
Yes, it's also equal to $\displaystyle (\cos \alpha + i \sin \alpha)^2$ by de Moivre's. I don't know if that helps though. Expanding doesn't yield anything useful.
It could also help to play around with the RHS, $\displaystyle i \tan (\alpha - \beta)$. Note that this is a purely imaginary number. You can try to reduce the LHS as $\displaystyle i \frac{\sin (\alpha - \beta)}{\cos (\alpha - \beta)}$ or something like that. You'll most probably need your trigonometric identities.
Yeah, it's an interesting problem. I haven't solved it yet...there seem to be several possible ways to the solution though. Just try a bunch of methods, see if one of them works. Maybe another user has a nice, elegant solution that doesn't require a lot of brute force (I don't want to brute force algebra/trig identities right now...). There might even be a geometric solution involving two points on a circle and finding the tangent of the angle in between.
let's write, for the time being, cos(2α) = u, sin(2α) = v, cos(2β) = y, sin(2β) = z.
then:
$\displaystyle \frac{a-b}{a+b} = \frac{u-y + i(v-z)}{u+y + i(v+z)} = \frac{((u-y) + i(v-z))((u+y) - i(v+z))}{(u+y)^2 + (v+z)^2}$
$\displaystyle = \frac{(u-y)(u+y) + (v-z)(v+z) + i(-(u-y)(v+z) + (v-z)(u+y))}{u^2 + 2uy + y^2 + v^2 + 2vz + z^2}$
$\displaystyle =\frac{u^2 + v^2 - y^2 - z^2 + i(2(vy - uz))}{u^2 + v^2 + y^2 + z^2 + 2(uy + vz)}$
now we can use some trig to simplify.
note that u^{2}+v^{2} = y^{2}+z^{2} = 1. so our messy fraction becomes:
$\displaystyle \frac{1 - 1 + i(2(vy - uz))}{2 + 2(uy + vz))} = i\frac{vy - uz}{1 + uy + vz}$
now $\displaystyle vy - uz = \sin(2\alpha)\cos(2\beta) - \cos(2\alpha)\sin(2\beta) = \sin(2\alpha - 2\beta)$
and $\displaystyle uy + vz = \cos(2\alpha)\cos(2\beta) + \sin(2\alpha)\sin(2\beta) = \cos(2\alpha - 2\beta)$.
so what we have is:
$\displaystyle i\frac{\sin(2\alpha - 2\beta)}{1 + \cos(2\alpha - 2\beta)} = i\tan\left(\frac{2\alpha - 2\beta}{2}\right) = i\tan(\alpha - \beta)$
Man ! Thank so very much Deveno.
You made it so simple to understand mate. You can't imagine the time i've spent on this problem.
Thank You again for your help.
Have a nice day !
edit: Thank you to Richard1234 too. I have solved it by your way too. Now I have two solutions.
Hello, ok59!
Out of curiosity, I brute-forced it . . . whew!
We need the Sum-to-Product Identities:
. . $\displaystyle \sin\!A\;+\;\sin\!B \:=\: 2\sin\!\left(\tfrac{A+B}{2}\right)\cos\!\left( \tfrac{A-B}{2}\right) \qquad \cos\!A\;+\;\cos\!B \:=\: 2\cos\!\left(\tfrac{A+B}{2}\right)\cos\!\left( \tfrac{A-B}{2}\right)$
. . $\displaystyle \sin\!A\;-\;\sin\!B \:=\:2\cos\!\left(\tfrac{A+B}{2}\right)\sin\!\left ( \tfrac{A-B}{2}\right) \qquad \cos\!A\;-\;\cos\!B \:=\:\text{-}2\sin\!\left(\tfrac{A+B}{2}\right)\sin\!\left( \tfrac{A-B}{2}\right) $
$\displaystyle \text{Given: }\:\begin{Bmatrix}a &=& \cos2A + i\sin2A \\ b &=& \cos2B + i\sin2B \end{Bmatrix}$
$\displaystyle \text{Prove that: }\:\frac{a-b}{a+b} \:=\; i\tan(A-B)$
$\displaystyle \text{The numerator is: }\:a - b \:=\:(\cos2A-\cos2B) + i(\sin2A - \sin 2B) $
x x . . . . . . . . . . . . . . . $\displaystyle =\;\text{-}2\sin(A\!+\!B)\sin(A\!-\!B) + 2i\cos(A\!+\!B)\sin(A\!-\!B) $
x x . . . . . . . . . . . . . . . $\displaystyle =\;2\sin(A\!-\!B)\bigg[\text{-}\sin(A\!+\!B) + i\cos(A\!+\!B)\bigg] $
$\displaystyle \text{The denominator is: }\:a + b \:=\:(\cos2A + \cos2B) + i(\sin2A +\sin2B)$
. . . . . . . . . . . . . . . . . . . $\displaystyle =\;2\cos(A\!+\!B)\cos(A\!-\!B) + 2i\sin(A\!+\!B)\cos(A\!-\!B)$
. . . . . . . . . . . . . . . . . . . $\displaystyle =\;2\cos(A\!-\!B)\bigg[\cos(A\!+\!B) + i\sin(A\!+\!B)\bigg]$
$\displaystyle \text{And we have: }\:\frac{a-b}{a+b} \;=\;\frac{\sin(A\!-\!B)\bigg[\text{-}\sin(A\!+\!B) + i\cos(A\!+\!B)\bigg]}{\cos(A\!-\!B)\bigg[\cos(A\!+\!B) + i\sin(A\!+\!B)\bigg]} $
. . . . . . . . . . . . . . . $\displaystyle =\;\tan(A\!-\!B)\left[\frac{\text{-}\sin(A\!+\!B) + i\cos(A\!+\!B)}{\cos(A\!+\!B) + i\sin(A\!+\!B)\right]} $
$\displaystyle \text{Rationalize: }\:\tan(A-B)\cdot\left[\frac{\text{-}\sin(A\!+\!B) + i\cos(A\!+\!B)}{\cos(A\!+\!B) + i\sin(A\!+\!B)}\right]\cdot\left[\frac{\cos(A\!+\!B) - i\sin(A\!+\!B)}{\cos(A\!+\!B) - i\sin(A\!+\!B)}\right] $
. . . . . . .$\displaystyle =\;\tan(A-B)\cdot \left[\frac{\text{-}\sin(A\!+\!B)\cos(A\!+\!B) + i\sin^2(A\!+\!B) + i\cos^2(A\!+\!B) - i^2\sin(A\!+\!B)\cos(A\!+\!B)} {\cos^2(A\!+\!B) - i^2\sin^2(A\!+\!B)}\right] $
. . . . . . .$\displaystyle =\;\tan(A-B)\cdot\left[\frac{\text{-}\sin(A\!+\!B)\cos(A\!+\!B) + i\sin^2(A\!+\!B) + i\cos^2(A\!+\!B) + \sin(A\!+\!B)\cos(A\!+\!B)}{\cos^2(A\!+\!B) + \sin^2(A\!+\!B)}\right] $
. . . . . . .$\displaystyle =\;\tan(A-B)\cdot\left[\frac{i\sin^2(A+B) + i\cos^2(A+B)}{1}\right] $
. . . . . . .$\displaystyle =\;\tan(A-B)\cdot i\left[\sin^2(A+B) + \cos^2(A+B)\right] $
. . . . . . .$\displaystyle =\;\tan(A-B)\cdot i \cdot 1$
. . . . . . .$\displaystyle =\;i\tan(A-B)$
De Moivre's Theorem is a relatively simple formula for calculating powers of complex numbers.De Moivre's formula states that for any real number x and any integer n, (cosx + isinx)n = cos(nx) + isin(nx).(r cisθ)^n = r^n cis(nθ) here n is an integer.
free algebra word problems
Here's another (equivalent) way: I'll begin with a general calculation that I'll then apply to this specific problem.
Suppose $\displaystyle z \in \mathbb{C} \ni z \neq -1$, and $\displaystyle |z|^2 = 1$ (i.e. $\displaystyle z$ on the unit circle except the leftmost point).
So let $\displaystyle z = e^{\theta i}$, where $\displaystyle \theta$ is not an odd multiple of $\displaystyle \pi$.
Then $\displaystyle \frac{z-1}{z+1}$
$\displaystyle = \frac{(z-1)(\bar{z}+1)}{(z+1)(\bar{z}+1)} = \frac{|z|^2+z-\bar{z}-1}{|z|^2+z+\bar{z}+1} = \frac{1+z-\bar{z}-1}{1+z+\bar{z}+1}$
$\displaystyle = \frac{z-\bar{z}}{2+(z+\bar{z})} = \frac{2Im(z)i}{2+2Re(z)} = \frac{Im(z)}{1+Re(z)}i = \frac{\sin(\theta)}{1+\cos(\theta)}i = \tan(\theta / 2)i$.
With that in hand, the result follows quickly. Let $\displaystyle a=e^{2\alpha i}, b=e^{2\beta i}$, $\displaystyle a+b \neq 0$. Let $\displaystyle z = a/b$. (Note that $\displaystyle |b| = 1$, so $\displaystyle b \neq 0$.)
Let $\displaystyle \theta = 2(\alpha - \beta)$. Then $\displaystyle z = e^{\theta i}$, and obviously $\displaystyle |z| = 1$.
Notice that from $\displaystyle a+b \neq 0$, it follows that $\displaystyle a \neq -b$, so that $\displaystyle z = a/b \neq -1$, and hence so can apply the above derivation with this $\displaystyle z$.
Now consider $\displaystyle \frac{a-b}{a+b}$. Have:
$\displaystyle \frac{a-b}{a+b} = \frac{a/b - 1}{a/b + 1} = \frac{z-1}{z+1} = \tan(\theta / 2)i = \tan((2(\alpha - \beta)) / 2)i = \tan(\alpha - \beta)i$ as desired.
Thus the claim is established.
Notice that all the "bad" cases amounted to $\displaystyle a + b = 0$, because then $\displaystyle a = -b = e^{(2n+1) \pi i}b$, so $\displaystyle e^{2\alpha i} = e^{(2n+1) \pi i}e^{2\beta i}$, so $\displaystyle \theta = 2(\alpha - \beta) = (2n+1) \pi$.
That would've made $\displaystyle z = e^{\theta i} = e^{(2n+1) \pi i} = -1$, and also would've maded $\displaystyle \alpha - \beta = (n+1/2) \pi$, which is where $\displaystyle \tan(\alpha - \beta)i$ is undefined.