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.
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.
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
$\displaystyle \chi$ $\displaystyle \sigma$
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 $\displaystyle \displaystyle \cos{x}$?
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.