How do you use De Moivre's Theorem to Prove this Identity?

Question

Use De Moivre's Theorem to prove:

let t = theta

sin2t = 2sin(t)cos(t)

I'm having trouble with this; I thought De Moivre only apply to complex numbers. How do you prove this Trigonometry identity.

Thanks for your help!

Answer 1

According De Moivre's theorem



cos(2t) + isin(2t) = [cos(t) + isin(t)]^2



expand RHS



[cos(t) + isin(t)]^2 = cos^2(t) + 2isin(t)cos(t) + i^2 sin^2(t)



=>cos^2(t) - sin^2(t) + i2sin(t) cos(t)



so cos(2t) + isin(2t) = cos^2(t) - sin^2(t) + i2sin(t) cos(t)



so comparing real and imaginary parts of LHS and RHS



cos(2t) = cos^2(t) - sin^2(t) and



isin(2t) = i2sin(t)cos(tx)



so sin(2t) = 2sin(t)cos(t)

Answer 2

De Moivre Theorem Proof

Answer 3

[ e^(2it) - e^(-2it) ] / 2i = sin (2t)

sin t = [ e^(it) - e^(-it) ] / 2i

cos t = [ e^(it) + e^(-it) ] / 2



On the RHS of the equation below, I skipped a step, which is canceling the denominators. This is just to let you know how I got it.



[ e^(2it) - e^(-2it) ] / 2i = [ e^(it) + e^(-it) ] * [ e^(it) - e^(-it) ] / 2i

e^(2it) - e^(-2it) = [ e^(it) + e^(-it) ] * [ e^(it) - e^(-it) ]



FOIL



e^(2it) - e^(-2it) = e^(2it) - e^(-2it)



You may not have seen what I done quite clearly.

Using De Moivre's Thereom we find what sin t, sin 2t and cos t is in terms of imaginary numbers. Then we manipulate using algebra.

-----

By the way, one of the steps looks like the following:



a^2 - b^2 = (a - b)(a + b)



We didn't have to FOIL, but in a proof I suggest we should, even if we realize this.

Answer 4

De Moivre's Theorem states



(cos(x)+isin(x))^n = cos(nx) + isin(nx)



let n = 2



(cos(x)+isin(x))^2 = cos(2x) + isin(2x)

isin(2x) = (cos(x)+isin(x))^2 - cos(2x)

isin(2x) = cos^2(x)+2icos(x)sin(x)-sin^2(x) - cos(2x)



now multiple i throughout



i^2(sin(2x)) = icos^2(x)+2(i^2)cos(x)sin(x)-isin^2(x) - icos(2x)

-sin(2x) = -2cos(x)sin(x) + i(cos^2(x)-sin^2(x)-cos(2x))

sin(2x) = 2cos(x)sin(x) - i(cos^2(x)-sin^2(x)-cos(2x))



cos(2x) = cos^2(x)-sin^2(x)



so the term i(cos^2(x)-sin^2(x)-cos(2x)) becomes zero



therefore,



sin(2x) = 2cos(x)sin(x)

Answer 5

all of us recognize t for theta e^it = cos t + i sin t sq. both area e^2it = (cos t + i sin t)^2 = cos^2 t + 2i cos t sin t - sin ^2 t cos 2t + i sin 2t = cos^2 t - sin ^2 t + 2i sin t cos t equting authentic area we get cos 2t = cos^2 t- sin ^2 t