Why $\cos x = 1 - \frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots$

We know $\sin x = \frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$ Derivation of both sides: $\cos x = \frac{1}{1!}-3\frac{x^2}{3!}+5\frac{x^4}{5!}-\cdots$ $\cos x = 1 - \frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots$ What a short and concise proof!

#Algebra

Easy Math Editor

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Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Why \(\cos x = 1 - \frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots\)

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