Let sin x and cos x be defined by their power series expansions

how can we use term by term differentiation to show that

$\displaystyle d/dx$sin x = cos x and

$\displaystyle d/dx$cos x = − sin x.

- Feb 17th 2011, 04:58 AMmaximus101Power series expansion proofs for the differential of sinx and cosx
Let sin x and cos x be defined by their power series expansions

how can we use term by term differentiation to show that

$\displaystyle d/dx$sin x = cos x and

$\displaystyle d/dx$cos x = − sin x. - Feb 17th 2011, 05:35 AMchisigma
Is...

$\displaystyle \displaystyle \sin x = \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!}$ (1)

... so that...

$\displaystyle \displaystyle \frac{d}{dx} \sin x = \sum_{n=0}^{\infty} (-1)^{n} (2n+1)\ \frac{x^{2n}}{(2n+1)!}= $

$\displaystyle \displaystyle = \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n}}{(2n)!} = \cos x$ (2)

The same procedure is valid for $\displaystyle \cos x$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Feb 17th 2011, 06:24 AMProve It
- Feb 17th 2011, 08:21 AMKrizalid
No it's not circular.

Note that he differentiate the power series (given hypothesis) and identified that series as cosine. - Feb 17th 2011, 08:28 AMProve It
The first sentence of the question is "Let $\displaystyle \displaystyle \sin{x}$ and $\displaystyle \displaystyle \cos{x}$ be defined by their power series expansions..."

How is one supposed to generate the power series expansions without already knowing the derivatives of $\displaystyle \displaystyle \sin{x}$ and $\displaystyle \displaystyle \cos{x}$? - Feb 17th 2011, 08:48 AMSammyS
Exactly the point!

Let sin x and cos x be**defined**by their power series expansions …

So**IF**you were to**define**the sine & cosine functions by what we know their power series are. Guess what!! No surprise that $\displaystyle \displaystyle {{d}\over{dx}}\,\sin(x)=\cos(x)\quad \text{ and }\quad {{d}\over{dx}}\,\cos(x)=-\sin(x)$

It's not so much circular reasoning as it is an exercise to help students verify that the power series have the same properties as the functions they represent - indeed they are identical.