Proving standard properties of sine and cosine defined by their power series

Question

Definition: We define $\displaystyle \sin x = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{\left ( 2n+1 \right )!}, \; x \in \mathbb{R} $ and $ \displaystyle \cos x = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{\left ( 2n \right )!}, \; x \in \mathbb{R}$.

Well I want to prove using these definitions that:

• $\displaystyle \sin^2 x+\cos^2 x =1 $
• $\displaystyle \left ( \sin x \right )'=\cos x, \; (\cos x)' =-\sin x$

I was able to prove most of the properties that these functions have using these definitions but not those two.

For the first one I started as follows: $$\begin{aligned} \sin^2 x +\cos^2 x &=\left ( \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{\left ( 2n+1 \right )!} \right )^2+\left ( \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{\left ( 2n \right )!} \right )^2 \\ &= \sum_{m=0}^{\infty}\sum_{n=0}^{\infty}\frac{\left ( -1 \right )^{m}x^{2m+1}}{\left ( 2m+1 \right )!}\frac{\left ( -1 \right )^{n}x^{2n+1}}{\left ( 2n+1 \right )!}+ \sum_{m=0}^{\infty}\sum_{n=0}^{\infty}\frac{\left ( -1 \right )^m x^{2m}}{\left ( 2m \right )!}\frac{(-1)^n x^{2n}}{\left ( 2n \right )!} \end{aligned}$$

then I don't know how to proceed .

As for the second one despite I differentiated the series , I cannot get the series of cosine and vice versa the minus series of sine.

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I also have another question regarding the definitions.

The classic definition of the trigonometric functions involves the unit circle. Suppose we define $\sin x, \; \cos x$ as above. Are these two definitions equivelant? Similary if we define $\displaystyle \sin x = \frac{e^{ix}-e^{-ix}}{2i}, \; \cos x =\frac{e^{ix}+e^{-ix}}{2}$ are these three definitions equivelant?

Answer 1

Justify that you can differentiate the series term by term and find, for all $x\in \mathbb R$, $$\sin'(x)=\sum \limits_{n=0}^\infty\left(\dfrac{(-1)^nx^{2n}}{(2n)!}\right)=\cos(x).$$

The first equality is straightforward, I can't imagine what you're missing. Let me know if you need it and I'll try to clarify.

Similarly $\cos'=-\sin$.

To prove that $\forall x\left((\sin(x))^2+(\cos(x))^2=1\right)$, consider the function $x\mapsto (\sin(x))^2+(\cos(x))^2$, differentiate, etc, etc.

"The classic definition of the trigonometric functions involves the unit circle. Suppose we define $\sin x, \; \cos x$ as above. Are these two definitions equivalent?"

Yes, if one accepts the geometric definitions as something meaningful, then they are equivalent.

"Similary if we define $\displaystyle \sin x = \frac{e^{ix}-e^{-ix}}{2i}, \; \cos x =\frac{e^{ix}+e^{-ix}}{2}$ are these three definitions equivalent?"

Note that $$e^{ix}+e^{-ix}=\sum \limits_{n=0}^\infty\left(\dfrac{i^nx^n}{n!}\right)+\sum \limits_{n=0}^\infty\left(\dfrac{(-1)^ni^nx^n}{n!}\right).$$

Now consider the cases $n\equiv k\pmod 4$ with $k\in \{0,1,2,3\}$.

It's not hard to see that some stuff will cancel (the imaginary parts, that is, when $n\equiv \pm 1\pmod 4 $) and others will add up (double up actually) and the definition of $\cos$ given at the top of the question appears.

Hence they are all equivalent.