Find out the answer to "Are there any other ways to demonstrate thatsin⁡(x)=∑k=0(−1)kx1+2k(1+2k)!"

e3r6a2n1dz60

Answered question

2022-03-15

Are there any other ways to demonstrate that

\sin (x) = \sum_{k = 0} \frac{{(- 1)}^{k} x^{1 + 2 k}}{(1 + 2 k)!}

InnovaRat0r5

Beginner2022-03-16Added 2 answers

There's the way Euler did it. First recall that

\sin (θ_{1} + θ_{2} + θ_{3} + \dots) = \sum_{o d d k \geq 1} {(- 1)}^{\frac{k - 1}{2}} \sum_{| A | = k} \prod_{i \in A} {\sin θ}_{i} \prod_{i \neq\in A} {\cos θ}_{i}

Then let n be an infinitely large integer and let

x = \frac{θ}{n} + \dots + \frac{θ}{n}

and apply the formula to find

\sin x

. Finally, recall that (as Euler would put it), since

\frac{θ}{n}

is infinitely small,

\sin (\frac{θ}{n}) = \frac{θ}{n}

and

\cos (\frac{θ}{n}) = 1

. Then do a bit of algebra and the series drops out.
The algebra will include things like saying that

\frac{n (n - 1) (n - 2) \dots (n - k + 1)}{n^{k} + 1} = 1

if n is an infinite integer and k is finite integer

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