A pretty easy one just for kicks, and it's a fairly standard proof.

 

Prove that [math]e^{i\theta} = \cos(\theta) + i\sin(\theta)[/math] and hence show [math]e^{\pi i} = -1[/math]

 

(for the more adventurous, try showing [math](\cos(\theta) + i \sin(\theta))^n = \cos(n\theta) + i\sin(n\theta)[/math]).

Just to give you a little hint and since I've had a question about what you can and can't assume;

 

You may assume standard results such as Taylor series of functions, for example.

 

Basically if you write a proof with standard results (within reason), you'll get the credit. Just do what you think is best

dont know if this constitutes a rigorous proof, but here we go anyway

 

Let us consider the function [math]y=\cos(x)+i\sin(x)[/math]

 

Continuing to treat [math]i[/math] like any other number, we have, by differentiation,

 

[math]\frac{dy}{dx}=-\sin(x)+i\cos(x)=i(\cos(x)+i\sin(x))[/math]

Hence,

[math]\frac{dy}{dx}=iy[/math]

[math]i\frac{dx}{dy}=\frac{1}{y}[/math]

[math]ix=\log(y)+c[/math]

 

But when x= 0 , y = 1 , So c = 0

[math]ix=\log(y)[/math]

[math]y=e^{ix}[/math]

 

Therefore [math]y=\cos(x)+i\sin(x)=e^{ix}[/math]

 

Another way to "show" this, is to use taylor expansions of /cosx and /sinx, form the linear combination [math]\cos(x)+i\sin(x)[/math] and show that its equivalent to the taylor expansion of [math]e^{ix}[/math], where [math]i[/math] is taken to be just another number.

 

For the second part of the question, you just plug in x=pi in the identity.

Those are two well know formulae. First is Euler's Formula and the second is de Moivre's Theorem if I'm not mistaken.

By using Taylor series, Euler's formula is easily proven.

And I may have seen the proof of de Moivre's theorem already, should leave it to some of you guys.

For de Moivre's, use the Euler's formula to write

[math]

(\cos x + i\sin x)^n = (e^{ix} )^n

[/math]

[math]

= e^{inx } = e^{i(\alpha )}

[/math]

where [math]\alpha=nx[/math]

Using Euler's Formula again,

[math]

=\cos\alpha+i\sin\alpha

[/math]

[math]

=\cos(nx)+i\sin(nx)

[/math]

Hurrah, we have a winner.

 

I'll try and make the problem slightly more interesting next time, was kinda busy so came up with the problem in a rush.

You can use e^(ix) = cos x + i sin x to prove many trig identities.

 

 

for example for the double angle formula

 

e^(i2x) = cos 2x + i sin2x (1)

 

but e^(i2x) = e^(ix) * e^(ix) = (cosx + isinx) * ( cosx + isinx) =

 

(cosx)^2 - (sinx)^2 + 2iSinxCosx (2)

 

so setting (1) equal to (2)

 

cos2x + isin2x = (cosx)^2 - (sinx)^2 + 2isinx cosx

 

equating both the real and imaginary parts we get

 

cos 2x = (cosx) ^2 - (sinx)^2

 

and

 

sin2x = 2sinx cosx