Maclaurin Series for the Sine Function
The Maclaurin series expansion of the sine function is: $$ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots + \frac{(-1)^n x^{2n+1}}{(2n+1)!} + o(x^{2n+1}) $$
Proof
We begin with the Maclaurin series formula:
$$ f(x) = \sum_{k=0}^{n} \frac{f^{(k)}(0)}{k!} \cdot x^k + o(x^n) $$
In this case, we have f(x) = sin(x).
For k = 0
Since f(0)(x) = f(x) = sin(x):
$$ \frac{f^{(0)}(0)}{0!} \cdot x^0 = \frac{\sin(0)}{0!} \cdot x^0 = \frac{0}{1} \cdot 1 = 0 $$
Thus, the first term in the Maclaurin series is zero:
$$ \sin(x) = 0 + o(x^0) = 0 + o(1) $$
For k = 1
The first derivative of sin(x) is cos(x):
$$ \frac{f^{(1)}(0)}{1!} \cdot x^1 = \frac{D[\sin(0)]}{1!} \cdot x = \frac{\cos(0)}{1} \cdot x = x $$
Therefore, the Maclaurin series up to the first-order term is:
$$ \sin(x) = 0 + x + o(x^1) $$
For k = 2
The second derivative of sin(x) is -sin(x):
$$ \frac{f^{(2)}(0)}{2!} \cdot x^2 = \frac{D^{(2)}\sin(0)}{2!} \cdot x^2 = \frac{-\sin(0)}{2!} \cdot x^2 = 0 $$
Hence, the second-order term in the Maclaurin series is zero:
$$ \sin(x) = 0 + x + 0 + o(x^2) $$
For k = 3
The third derivative of sin(x) is -cos(x):
$$ \frac{f^{(3)}(0)}{3!} \cdot x^3 = \frac{D^{(3)}\sin(0)}{3!} \cdot x^3 = \frac{-\cos(0)}{3!} \cdot x^3 = -\frac{x^3}{3!} $$
Thus, the third-order term in the Maclaurin series is:
$$ \sin(x) = 0 + x + 0 - \frac{x^3}{3!} + o(x^3) $$
For k = 4
The fourth derivative of sin(x) is sin(x):
$$ \frac{f^{(4)}(0)}{4!} \cdot x^4 = \frac{D^{(4)}\sin(0)}{4!} \cdot x^4 = \frac{\sin(0)}{4!} \cdot x^4 = 0 $$
Therefore, the fourth-order term in the Maclaurin series is zero:
$$ \sin(x) = 0 + x + 0 - \frac{x^3}{3!} + 0 + o(x^4) $$
For k = 5
The fifth derivative of sin(x) is cos(x):
$$ \frac{f^{(5)}(0)}{5!} \cdot x^5 = \frac{D^{(5)}\sin(0)}{5!} \cdot x^5 = \frac{\cos(0)}{5!} \cdot x^5 = \frac{x^5}{5!} $$
So the fifth-order term in the Maclaurin series is:
$$ \sin(x) = 0 + x + 0 - \frac{x^3}{3!} + 0 + \frac{x^5}{5!} + o(x^5) $$
For k = 6
We observe that the terms alternate between nonzero and zero values.
Hence, for k = 6, the Maclaurin series remains the same as for k = 5, except for the higher-order term o(x6):
$$ \sin(x) = 0 + x + 0 - \frac{x^3}{3!} + 0 + \frac{x^5}{5!} + 0 + o(x^6) $$
Therefore, the Maclaurin series for sin(x) up to the sixth-order term is:
$$ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} + o(x^6) $$
Graphically, it looks like this:
And so on.
