I know that $$\sin(2x)= 2x - \frac{8x^3}{3!} + \frac{32x^5}{5!} - \frac{128x^7}{7!} + \cdots $$ $$\sin(2x)= \sum_{n=0}^\infty (-1)^n {2^{2n+1}x^{2n+1} \over (2n+1)!}$$ But I don't see how I can use that series to find the series for $\cos(2x)$? Is there any way I can use the Trig Identity $\cos(2x)=1-2\sin(x)^2$ ?
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$\begingroup$For a positive $R$, this series expansion $$\sin(2x)= \sum_{n=0}^\infty (-1)^n {2^{2n+1}x^{2n+1} \over (2n+1)!}$$ is valid for $x$ in $[-R, R]$. Differentiating term by term gives $$2\cos(2x) = \sum_{n=0}^\infty (-1)^n {2^{2n+1}x^{2n} \over (2n)!},$$ valid for $x$ in $(-R, R)$. Hence this series expansion for $\cos(2x)$ is valid over the whole real line because $R$ was arbitrary.
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