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De Moivre's Theorem

Please help me , i am really annoyed by this problem for 2 days. if a = cos 2alpha + i sin 2alpha b = cos 2beta + i sin 2beta prove that, a - b ______ = i tan (alpha - beta ) a + b does ( cos alpha + i sin alpha )^2 has any advantage here ? how do I supposed to get "tan" from this ?

(edited 11 years ago)

Reply 1

ghostwalker

I'd start by multiplying top and bottom by the conjugate of the denominator, and use that fact that:

z\bar{z}=|z|^2 =1

, in this case (where z is a or b), before you do any expansions in terms of sine and cosine.

Your tan will arise from dividing top and bottom by cosine as some stage - not worked it through.

You also know that in general,

z+\bar{z}= 2\text{Re}(z)

and

z-\bar{z}= 2i\text{Im}(z)

, which will also be useful, I think.

(edited 11 years ago)

Reply 2

username667907

Original post by ok5

Please help me , i am really annoyed by this problem for 2 days.

if a = cos 2alpha + i sin 2alpha
b = cos 2beta + i sin 2beta

prove that,

a - b
______ = i tan (alpha - beta )
a + b

does ( cos alpha + i sin alpha )^2 has any advantage here ?
how do I supposed to get "tan" from this ?

you're probably going to get the tan by getting a

\dfrac{sin\theta}{cos\theta}

i'm attempting to solve this now... not sure how close I'll be able to get though.

Edit: i've done it. Follow the advice of the above poster. If you get stuck i can post the solution wrapped in many layers of spoilers.

(edited 11 years ago)

Reply 3

nuodai

There's quite a clean way that you can do it without having to multiply out brackets. Notice that

2\alpha = (\alpha + \beta) + (\alpha - \beta)

and

2\beta = (\alpha + \beta) - (\alpha - \beta)

. So

e^{2i\alpha} = e^{i(\alpha + \beta)}e^{i(\alpha - \beta)}

and

e^{2i\beta} = e^{i(\alpha + \beta)}e^{-i(\alpha - \beta)}

. Notice the common factor of

e^{i(\alpha+\beta)}

. So you can take out this from both the numerator and denominator, and hence cancel it, and what you're left with is something you can easily recognise as being

i\tan(\alpha - \beta)

.

(This assumes that you've already used de Moivre's theorem to write out the fraction in terms of exponentials... if you haven't, then do that.)

(edited 11 years ago)

Reply 4

username667907

I decided to start writing out my solution. This probably isn't the quickest way.

Notes:

cos\theta + i sin\theta = e^{i\theta}

e^{0} = 1

sin2\theta = 2 sin\theta cos\theta

cos2\theta = cos^2\theta - sin^2\theta

sin^2\theta + cos^2\theta = 1

Throughout this example i have said

\alpha - \beta = X

for convenience.

Spoiler

Reply 5

ok5

Thank you guys you all were really helpful. I learned something new today. Thanks to phredd for elaborating it nicely and nuodai and ghostwalker for showing me the alternate way to do it. Thanks again to all of you. Have a nice day ! :smile:

Reply 6

ghostwalker

Original post by Phredd

...

+rep for the deepest meaningful level of spoilers I've ever seen.

Would rep nuodai too, but have to make up my quota of 40 other people first (or however many it is).

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