Find (cos(4t) + i sin(4t))^3.?

Question

I've already tried distribution, I want to know if there are any other solutions.

Answer 1

cos4t + i sin4t = e^i4t
(cos4t + i sin4t)^3 = (e^i4t)^3
(cos4t + i sin4t)^3 = e^i12t
(cos4t + i sin4t)^3 = cos 12t + i sin12t

Answer 2

Remember, by definition: cis θ = cos θ + i · sin θ

So, cos (4t) + i·sin (4t) = cis(4t)

And, therefore, [ cos (4t) + i·sin (4t) ]³ = [ cis(4t) ]³

==========

Using DeMoivres Theorem:
(r · cis θ)^n = r^n · cis (n·θ)

[ 1 · cis(4t) ]³ =
1³ · cis(3 · 4t) =
1 · cis(12t) =
cis(12t)


===========

Using Eulers Theorem:
cis θ = e^(i·θ)

cis(4t) = e^(i·4t)

[ cis(4t) ]³ = [ e^(i·4t) ]³

By rules of exponents:
[ e^(i·4t) ]³ =
e^(3·i·4t) =
e^(i·12t)

And by Eulers Theorem again,
e^(i·12t) =
cis(12t)


===========

Either way... you end up with cis(12t)

By definition,
cis(12t) = cos(12t) + i · sin(12t)

Answer 3

Use Euler's forumula: e^(iθ) = cos(θ) + isin(θ). THEN carry though the exponent.

[ cos(4t) + i sin(4t) ]^3
[ e^(i * 4t) ] ^3
e^(i * 12t)
cos(12t) + i sin(12t)

Answer 4

cos4t +isin 4t = (cost + isint)^4
So (cos4t +isin 4t)^3 = (cost + isint)^12

Answer 5

Use deMoivre's formula: [cos (x) + i sin (x)]^n = cos(nx) + i sin(nx). ( http://en.wikipedia.org/wiki/Demoivre%27s_theorem )

Therefore, (cos(4t) + i sin(4t))^3 = cos (12t) + i sin(12t).

Answer 6

[exp(4it)]^3 = exp(12it)
= cos(12t) + i*sin(12t)

Answer 7

exp(i12t)