Find the Maclaurin expansion for \(\sin ^{2} x\). (Hint: use a trigonometrical identity and the series for \(\sin x\).)

The Maclaurin series for the sine function is a widely known result, which can be written as: $$\sin x = \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$$

Now, we must use the double angle formula to write \(\sin^2 x\) in terms of trigonometric functions, using the identity: $$\sin^2 x = \frac{1}{2} - \frac{1}{2} \cos(2x)$$

Next, we need to find the Maclaurin series for \(\cos(2x)\). We know that the Maclaurin series for the cosine function is: $$\cos x = \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots$$ So, for \(\cos(2x)\), we can substitute \(2x\) in the above series: $$\cos(2x) = \sum_{n=0}^{\infty}(-1)^n\frac{(2x)^{2n}}{(2n)!} = 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \frac{(2x)^6}{6!} + \dots$$

Now we can find the Maclaurin expansion for \(\sin^2 x\) by replacing \(\cos(2x)\) with its expansion in the double angle formula we've derived in step 2: $$\sin^2 x = \frac{1}{2} - \frac{1}{2}\left(1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \frac{(2x)^6}{6!} + \dots\right)$$ Now, expand the terms and simplify the series: $$\sin^2 x = \frac{1}{2} - \frac{1}{2} + \frac{1}{2}\frac{(2x)^2}{2!} - \frac{1}{2}\frac{(2x)^4}{4!} + \frac{1}{2}\frac{(2x)^6}{6!} - \dots$$ $$\sin^2 x = \frac{x^2}{2} - \frac{2^2x^4}{2\cdot4!} + \frac{2^4x^6}{2\cdot6!} - \frac{2^6x^8}{2\cdot8!} + \dots$$ So, the Maclaurin expansion for \(\sin^2x\) is: $$\sin^2 x = \frac{x^2}{2} - \frac{x^4}{12} + \frac{2x^6}{180} - \frac{8x^8}{5040} + \dots$$