Problem Statement
Find the first three non-zero terms in the Maclaurin series expansion of
$$f(x) = \ln(1 + \sin x)$$
Verified Solution & Marking Scheme
Evaluate successive derivatives at x = 0
1. $f(0) = \ln(1 + \sin 0) = \ln 1 = 0$.
2. $f'(x) = \frac{\cos x}{1 + \sin x} \implies f'(0) = \frac{\cos 0}{1 + 0} = 1$.
3. $f''(x) = \frac{-\sin x(1 + \sin x) - \cos x(\cos x)}{(1 + \sin x)^2} = \frac{-\sin x - \sin^2 x - \cos^2 x}{(1 + \sin x)^2} = \frac{-(1 + \sin x)}{(1 + \sin x)^2} = \frac{-1}{1 + \sin x}$.
So $f''(0) = \frac{-1}{1 + 0} = -1$.
4. $f'''(x) = \frac{d}{dx}[-(1 + \sin x)^{-1}] = (1 + \sin x)^{-2} \cos x = \frac{\cos x}{(1 + \sin x)^2}$.
So $f'''(0) = \frac{\cos 0}{(1 + 0)^2} = 1$.
Form Maclaurin series using Taylor theorem
$$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots$$
$$f(x) = 0 + 1x + \frac{-1}{2}x^2 + \frac{1}{6}x^3 + \dots = x - \frac{1}{2}x^2 + \frac{1}{6}x^3$$
Alternatively, using standard expansions:
$\sin x = x - \frac{x^3}{6} + \dots$ and $\ln(1 + u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \dots$:
$$\ln(1 + \sin x) = \left(x - \frac{x^3}{6}\right) - \frac{1}{2}\left(x - \frac{x^3}{6}\right)^2 + \frac{1}{3}x^3 = x - \frac{x^3}{6} - \frac{x^2}{2} + \frac{x^3}{3} = x - \frac{x^2}{2} + \frac{x^3}{6}$$