Find a 10 terms Taylor's series approximation?

Question

Find a 10 terms Taylor's series approximation to f(x) = cos x and f(x) = sin x,

I have no idea how to do this problem, can anyone help me start?

Answer 1

f(x) = cos(x), f(0) = 1, a0 = 1/1
f'(x) = - sin(x), f'(0) = 0, a1 = 0/1
f2(x) = - cos(x), f2(0) = - 1, a2 = - 1/2!
f3(x) = sin(x), f3(0) = 0
f4(x) = cos(x), f4(0) = 1, a4 = 1/4!
f5(x) = - sin(x), f5(0) = 0
f6(x) = - cos(x), f6(0) = - 1, a6 = - 1/6!
f7(x) = sin(x), f7(0) = 0
With the pattern established,
f(x) = 1 - x^2/2! + x^4/4! - x^6/6! + x^8/8! - x^10/10! + x^12/12! - x^14/14! + x^16/16! - x^18/18!

for f(x) = sin(x)
f(x) = x/1! - x^3/3! + x^5/5! - x^7/7! + x^9/9! - x^11/11! + x^13/13! - x^15/15! + x^17/17! - x^19/19!

Answer 2

f(x) = cos x,
f'(x) = -sin x
f"(x) = -cos x
f"'(x) = sin x
f""(x) = cos x which equal to f(x).
and thus f'(x) = f""'(x) and so forth.

From Taylor Series,
f(x) = f(0)+ [ f'(0)/1! ] x + [ f''(0)/2! ] x^2 + ...
f(x) = 1 - (1/2!) x^2 + (1/4!) x^4 - (1/6!) x^6 + (1/8!) x^8 - (1/10!) x^10

f(x) = 1 - (1/2) x^2 + (1/24) x^4 - (1/720) x^6 + (1/40320) x^8 - (1/3628800) x^10

Repeat the same method of f(x) = sin x