# Maclaurin Series for Sine and Cosine

Recall that a Maclaurin series of a function $f$ is Taylor series centered about $x = 0$ and that has the form $\sum_{n=0}^{\infty} f^{(n)}(0) \frac{x^n}{n!} = f(0) + f'(0)\frac{x}{1!} + f''(0)\frac{x^2}{2!} + ...$. We will now verify the Maclaurin series for the sine function $f(x) = \sin x$.

First take a look at the following table containing the derivatives of $f$ and the derivatives of $f$ evaluated at $x = 0$:

$f(x) = \sin x$ | $f(0) = \sin (0) = 0$ |

$f'(x) = \cos x$ | $f'(0) = \cos (0) = 1$ |

$f''(x) = -\sin x$ | $f''(0) = -\sin (0) = 0$ |

$f'''(x) = -\cos x$ | $f'''(0) = -\cos(0) = -1$ |

$f^{(4)}(x) = \sin x$ | f^{(4)}(0) = \sin (0) = 0 $]] |

Applying this to the formula of a Maclaurin series we get that:

(1)Now let's find the radius of convergence of this Maclaurin series using the ratio test as follows:

(2)Therefore the radius of convergence is $R = \infty$ and so this series converges for all $x \in \mathbb{R}$.

We now need to show that this Maclaurin series converges to $\sin x$. We will do this utilizing differential equations. Let $g(x)$ be the sum of the Maclaurin series given above, that is:

(3)Notice that if we differentiate this equation twice we get that:

(4)Therefore $g(x)$ satisfies the differential equation $g(x) = -g''(x)$, and so $g(x) + g''(x) = 0$. The general solution to this equation is in the form:

(5)Notice that if $x = 0$, then $0 = g(0) = A$ and so $A = 0$. Furthermore if we differentiate both sides of this equation we get $g'(x) = -A \sin x + B \cos x$, and $g'(0) = 1 = B$ and so $B = 1$, therefore we have that:

(6)Therefore $\sin x = \sum_{n=0}^{\infty} \frac{(-1)^{n}x^{2n+1}}{(2n+1)!}$ for all $x \in \mathbb{R}$.

Now if we differentiate both sides of this equation we have that:

(7)This is satisfied for all $x \in \mathbb{R}$.