Home » Blog » Verify that sin2x = ∑ (-1)n+1 (22n-1 / (2n)!) x2n

Verify that sin2x = ∑ (-1)n+1 (22n-1 / (2n)!) x2n

Assuming that \sin^2 x has a power-series representation in terms of powers of x, verify that it has the form

    \[ \sin^2 x = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{2^{2n-1}}{(2n)!}x^{2n} \]

valid for all real x.

Following the hint, we recall the trig identity

    \[ \cos 2x  = 1 - 2 \sin^2 x \qquad \implies \qquad \sin^2 x = \frac{1}{2} \left( 1 - \cos (2x) \right). \]

Then, we know the expansion for \cos x is

    \[ \cos x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!}. \]

So, putting these together we have

    \begin{align*}  \sin^2 x &= \frac{1}{2} - \frac{1}{2} \cos(2x) \\[9pt]  &= \frac{1}{2} - \frac{1}{2} \left( \sum_{n=0}^{\infty} \frac{(-1)^n (2x)^{2n}}{(2n)!} \right)\\[9pt]  &= \frac{1}{2} + \frac{1}{2} \left( \sum_{n=0}^{\infty} \frac{(-1)^{n+1}(2x)^{2n}}{(2n)!} \right) \\[9pt]   &= \frac{1}{2} + \sum_{n=0}^{\infty} \frac{(-1)^{n+1} 2^{2n-1}}{(2n)!} x^{2n} \\[9pt]  &= \frac{1}{2} + \frac{-1}{2} + \sum_{n=1}^{\infty} \frac{(-1)^{n+1} 2^{2n-1}}{(2n)!} x^{2n} \\[9pt]  &= \sum_{n=1}^{\infty} \frac{(-1)^{n+1} 2^{2n-1}}{(2n)!} x^{2n}.  \end{align*}

This is valid for all x \in \mathbb{R} since the expansion for cosine is valid for all x \in \mathbb{R}.

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):