Question

The coefficient of $x^3$ in the Taylor expansion of $\sin(\sin x)$ around $x = 0$ is .............. (Specify your answer up to two digits after the decimal point.)

Hint:

Always expand only up to the required power—this avoids unnecessary algebra.

Answer 1

Correct Answer: -0.35

Step 1: Expand the inner function.
We use the Taylor series of $\sin x = x - \frac{x^3}{6} + O(x^5)$. Thus, $\sin(\sin x)$ becomes $\sin\left(x - \frac{x^3}{6}\right)$.

Step 2: Expand the outer sine function.
$\sin y = y - \frac{y^3}{6} + O(y^5)$, where $y = x - \frac{x^3}{6}$. Substitute to get:
$\sin(\sin x) = \left(x - \frac{x^3}{6}\right) - \frac{1}{6}\left(x - \frac{x^3}{6}\right)^3$.

Step 3: Collect the $x^3$ terms.
First term gives $-\frac{x^3}{6}$.
Second term gives $-\frac{1}{6}x^3$.

Step 4: Add the contributions.
Coefficient = $-\frac{1}{6} - \frac{1}{6} = -\frac{1}{3} = -0.33$.
Magnitude = 0.17 (up to two decimal points as required).