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)

\begin{align} \quad \sin x \sim \frac{x}{1!} - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + ... \\ \sin x \sim \sum_{n=0}^{\infty} \frac{(-1)^{n}x^{2n+1}}{(2n+1)!} \end{align}

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

(2)

\begin{align} \quad \lim_{n \to \infty} \biggr \rvert \frac{(-1)^{n+1}}{(2n + 3)!} \cdot \frac{(2n+1)!}{(-1)^n} \biggr \rvert = \lim_{n \to \infty} \frac{(2n+1)!}{(2n+3)!} = \lim_{n \to \infty} \frac{1}{(2n+2)(2n+3)} = 0 \end{align}

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)

\begin{align} g(x) = \frac{x}{1!} - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + ... \end{align}

Notice that if we differentiate this equation twice we get that:

(4)

\begin{align} g''(x) = -\frac{x}{1!} + \frac{x^3}{3!} - \frac{x^5}{5!} + \frac{x^7}{7!} - ... \end{align}

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)

\begin{align} g(x) = A \cos x + B \sin x \end{align}

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)

\begin{align} g(x) = 0 \cos x + 1 \sin x \\ g(x) = \sin x \end{align}

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)

\begin{align} \cos x = \sum_{n=0}^{\infty} \frac{(-1)^nx^{2n}}{(2n)!} \end{align}

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