To prove that \(\sqrt{1 + \sin(2x)} = 1 + x - \frac{x^2}{2} - \frac{x^3}{6} + \frac{x^4}{24}\) using the Maclaurin series, we proceed step by step:
Step 1: Expand \(\sin(2x)\) using its Maclaurin series
The Maclaurin series for \(\sin(x)\) is:
\[
\sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots
\]
Substituting \(2x\) for \(x\), we have:
\[
\sin(2x) = 2x - \frac{(2x)^3}{6} + \frac{(2x)^5}{120} - \cdots
\]
Simplify:
\[
\sin(2x) = 2x - \frac{8x^3}{6} + \frac{32x^5}{120} - \cdots
\]
\[
\sin(2x) = 2x - \frac{4x^3}{3} + \frac{4x^5}{15} - \cdots
\]
Step 2: Expand \(\sqrt{1 + y}\) using its Maclaurin series
The Maclaurin series for \(\sqrt{1 + y}\) is:
\[
\sqrt{1 + y} = 1 + \frac{y}{2} - \frac{y^2}{8} + \frac{y^3}{16} - \cdots
\]
Step 3: Substitute \(y = \sin(2x)\) into \(\sqrt{1 + y}\)
We substitute \(\sin(2x)\) into the series for \(\sqrt{1 + y}\). Using only terms up to \(x^4\) for accuracy:
\[
\sqrt{1 + \sin(2x)} = 1 + \frac{\sin(2x)}{2} - \frac{\sin^2(2x)}{8} + \cdots
\]
Step 4: Substitute the series for \(\sin(2x)\)
Substitute \(\sin(2x) = 2x - \frac{4x^3}{3} + \cdots\) into \(\sqrt{1 + \sin(2x)}\):
\[
\sqrt{1 + \sin(2x)} = 1 + \frac{1}{2}\left(2x - \frac{4x^3}{3} + \cdots\right) - \frac{1}{8}\left(2x - \frac{4x^3}{3} + \cdots\right)^2 + \cdots
\]
Simplify term by term:
1. First term: \(1\)
2. Second term: \(\frac{1}{2}(2x) = x\)
3. Third term: \(\frac{1}{2}\left(-\frac{4x^3}{3}\right) = -\frac{2x^3}{3}\)
4. Fourth term (from \(-\frac{\sin^2(2x)}{8}\)):
\[
\sin^2(2x) \approx (2x)^2 = 4x^2 \quad \text{(using only up to \(x^2\))}.
\]
Substituting: \(-\frac{1}{8}(4x^2) = -\frac{x^2}{2}\)
Thus:
\[
\sqrt{1 + \sin(2x)} = 1 + x - \frac{x^2}{2} - \frac{x^3}{6} + \frac{x^4}{24}
\]
This matches the given expression. Hence proved.
