Prove: sin θ = [ e^(iθ) - e^(-iθ) ] / 2i and cos θ = [ e^(iθ) + e^(-iθ) ] / 2?

Question

Prove:
cos θ = [ e^(iθ) + e^(-iθ) ] / 2
sin θ = [ e^(iθ) - e^(-iθ) ] / 2i

I already know that cis θ = e^(iθ)

And that:
cos θ = Re{ cis θ } = Re{ e^(iθ) }
sin θ = Im{ cis θ } = Im{ e^(iθ) }

But how do I actually extract the functions, in terms of e, of each trigonometric function?

How do I prove:
cos θ = [ e^(iθ) + e^(-iθ) ] / 2
sin θ = [ e^(iθ) - e^(-iθ) ] / 2i

I imagine that I need to use the Taylor series of each trig function?

Answer 1

Okay, following your idea of using the Maclaurin's Expansion:
(And so that there is no argument here: we will 'set' x^0 = 1. Mathematically, it should be separated from the series, but we will include it here to save space only.)

Now, e^x = Σ x^k / k! , k = 0,..., ∞
cos[x] = Σ (-1)^m x^(2m) / (2m)! , m = 0,..., ∞
sin[x] = Σ (-1)^m x^(2m+1) / (2m+1)! , m = 0,..., ∞
This is assuming you know how to obtain the Taylor's expansion about a=0.

Now, e^(iθ) = Σ (iθ)^k / k! , k = 0,..., ∞
= 1 + iθ - {θ^2}/2 - i{θ^3}/6 + {θ^4}/24 + i{θ^5}/5! - {θ^6}/6! + ... + {iθ}^k/k! + ...
And, e^(-iθ) = Σ (-iθ)^k / k! , k = 0,..., ∞
= 1 - iθ - {θ^2}/2 + i{θ^3}/6 + {θ^4}/24 - i{θ^5}/5! - {θ^6}/6! +... + {-iθ}^k/k! + ...

Thus e^(iθ) + e^(-iθ) =
2 - 2 {θ^2}/2 + 2{θ^4}/4! - 2{θ^6}/6! + ...

Thus [e^(iθ) + e^(-iθ)]/2 =
1 - {θ^2}/2 + {θ^4}/4! - {θ^6}/6! + ... + (-1)^m θ^(2m)/(2m)! + ... = cos[θ]

Meanwhile, e^(iθ) - e^(-iθ) =
2iθ - 2i{θ^3}/3! + 2i{θ^5}/5! - 2i{θ^7}/7! + ...
Thus, [e^(iθ) - e^(-iθ)]/2i =
θ - {θ^3}/3! + {θ^5}/5! - {θ^7}/7! + ... + (-1)^m θ^(2m+1)/(2m+1)! + .. = sin[θ].

There can still be other ways of showing this...


§
LOL... simpler approach...
e^(iθ) = cos[θ] + i sin[θ]
e^(-iθ) = cos[-θ] + i sin[-θ] = cos[θ] - i sin[θ]

Thus e^(iθ) + e^(-iθ) = 2 cos[θ]
cos[θ] = [e^(iθ) + e^(-iθ)] / 2

And e^(iθ) - e^(-iθ) = 2i sin[θ]
sin[θ] = [e^(iθ) - e^(-iθ)] / 2i.
qed

Answer 2

Do you know this : e^(a+ib)=cos a + i*sin b
so e^(a-ib)=cos a - i*sin b
e^(ib)+e^(-ib)=cos a + i*sin b + cos a - i*sin b = 2*cos a
so we have cos a = [ e^(ib) + e^(-ib) ]/ 2
now : e^(ib)-e^(-ib)=cos a + i*sin b - cos a + i*sin b = 2*i*sin b
so we have sin a = [ e^(ib) - e^(-ib) ]/ 2i