Using a Taylor's series or otherwise; derive Euler's Formula

Use the Taylor series expansion for the following three functions: f(θ) = e^(iθ), g(θ) = cos(θ) and h(θ) = sin(θ). We should find that f(θ) = e^(iθ) = 1 + iθ - (θ^2/2!) - i(θ^3/3!) + ... = Sum(θ^n/n!), g(θ) = cos(θ) = 1 - (θ^2/2!) + (θ^4/4!) - (θ^6/6!) + ... = Sum((-1)^n . (θ^2n))/2n!) and finally, g(θ) = sin(θ) = θ - (θ^3/3!) + (θ^5/5!) + ... = Sum((-1)^n . (θ^2n+1))/2n+1!). Now it is a case of manipulating a our results for our functions to match Euler's Formula. Since we know e^(iθ) = cos(θ) + isin(θ) is Euler's Formula, and that we've been asked to use a Taylor series expansion, it is just a case of algebraic manipulation, starting from either the LHS or the RHS to achieve the other part of the equation.Let's start from the LHS (for powers of θ up to 5) : e^(iθ) = 1 + iθ - (θ^2/2!) - i(θ^3/3!) + (θ^4/4!) + i(θ^5/5!) - ... = (1 - (θ^2/2!) + (θ^4/4!) - ...) + i(θ - (θ^3/3!) + (θ^5/5!) - ...) and so on. If you notice the first term corresponds to the Taylor expansion of cos(θ) and the second to the expansion of i(sin(θ)) and hence we can say that e^(iθ) = cos(θ) + isin(θ) and the derivation of Euler's Formula using a Taylor's series is complete.