證明sinx=x-x³/3!+x⁵/5!-x⁷/7!+...

We can use the Maclaurin's series (Taylor's series of expansion at x = 0), where:

f(x) = f(0) + x f'(0) + x2 f"(0)/2! + x3 f(3)(0)/3! + ...

= Σ (n = 0 → ∞) xn f(n)(0)/n!

Let f(x) = sin x, then:

f(0) = 0

f'(0) = 1

f"(0) = 0

f(3)(0) = -1

Then in general we have:

f(2n)(0) = 0 for any non-negative integer n.

f(2n-1)(0) = (-1)n-1 for any non-negative integer n.

Hence the Maclaurin's series becomes

sni x = Σ (n = 0 → ∞) xn f(n)(0)/n!

= Σ (n = 0 → ∞) x2n-1 f(2n-1)(0)/(2n - 1)!

= x - x3/3! + x5/5! - x7/7! + ...