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De Moivre's Theorem

Ben

How would I find tan 3x in terms of tan x?

Reply 1

El Matematico

Id use the formula:

tan(A+B) = (tanA + tanB)/(1 - tanAtanB)

Reply 2

generalebriety

El Matematico

Id use the formula:

tan(A+B) = (tanA + tanB)/(1 - tanAtanB)

Very helpful, considering the thread title is "De Moivre's theorem".

OP: we know that z = cos 3x + i sin 3x = (cos x + i sin x)^3. Expand that right hand side by the binomial expansion, and you get z = cos 3x + i sin 3x = a + ib for some functions a and b. Then you know that tan 3x = Im(z) / Re(z) from the first equality, so do the same to the last equality...

Reply 3

Hopping Mad Kangaroo

Break tan 3x into sin 3x over cos 3x. Then apply the theorem to both sin 3x and cos 3x (post again if you would like me to show you how...)

Reply 4

Ben

Ok, thanks.

I now have

\frac{tan x (3cos^2 x - sin^2 x)}{cos^2 x - 3sin^2 x}

, but I'm not sure where to go from there.

Reply 5

generalebriety

Divide through by something that'll get rid of cos^2 x, and turn sin^2 x into tan^2 x... :wink:

Reply 6

Ben

Heh, thanks. Not sure why I didn't do that before. :redface:

Reply 7

Ben

I have another problem now.

I need to express sin^4 x in terms of cosines of multiples of x, but I can't get it to work. :confused:

Reply 8

nota bene

I think that would be nicer to do if you used

cos(x)=\frac{e^{ix}+e^{-ix}}{2}

and

sin(x)=\frac{e^{ix}-e^{-ix}}{2i}

Reply 9

The Bachelor

I think a non-complex method is in order here:

\sin^4 x = (\sin^2 x)(\sin^2 x)

Spoiler

Reply 10

The Bachelor

If you were to do it the complex way (which, imo, is harder in this case), you would use DeMoivre's theorem:

Spoiler

Reply 11

Arjy

how do you do it if you split it into sin/cos?

Reply 12

Hopping Mad Kangaroo

Arjy

how do you do it if you split it into sin/cos?

Equate coefficients. We know that cos nx + i sin nx = (cos x + isinx)^n
Then we multiply out the right hand side
To get cos nx = ... we ignore the terms of the expansion with i in them
To get sin nx = ... we ignore the terms of the expansion without i in them.