Find the Maclaurin series of sin(x) up to and including the degree 7 term.

Step-by-step solution

1. 1

   Build the derivative cycle at $x=0$. The derivatives of sine repeat with period four, so the coefficients follow a four-step pattern:

   $\sin 0=0,\quad \cos 0=1,\quad -\sin 0=0,\quad -\cos 0=-1,\quad \text{then repeat}$

   so $f^{(n)}(0)$ cycles through $0,1,0,-1$.

2. 2

   Insert those values into the Maclaurin formula. With $f(x)=\sum_{n\ge 0}\dfrac{f^{(n)}(0)}{n!}x^n$, every even derivative vanishes and only odd powers survive:

   $\sin x=\frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots$

   The absence of even powers is not a coincidence — $\sin(-x)=-\sin x$ makes the function odd, and an odd function can only have odd-degree terms.

3. 3

   Evaluate the factorials.

   $3!=6,\qquad 5!=120,\qquad 7!=5040$

   $\sin x=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+O\!\left(x^9\right)$

4. 4

   Write the general term. In closed form,

   $\sin x=\sum_{k=0}^{\infty}\frac{(-1)^k x^{2k+1}}{(2k+1)!}$

   which converges for every real $x$ — the radius of convergence is infinite, because $(2k+1)!$ eventually outgrows any fixed power.

5. 5

   Measure the accuracy. At $x=1$ radian the truncation gives

   $1-\frac16+\frac{1}{120}-\frac{1}{5040}=0.8414682\ldots$

   against $\sin 1=0.8414710$, an error of about $2.7\times 10^{-6}$ — consistent with the next omitted term $\tfrac{x^9}{9!}=\tfrac{1}{362880}\approx 2.76\times 10^{-6}$. The alternating-series bound guarantees the error never exceeds that first dropped term.