Proof of $|\sin(x)| \leq 1$ and $|\cos(x)| \leq 1$

Question

Using the series representations of $\sin(x)$ and $\cos(x)$, how does one show that both $|\sin(x)| \leq 1$ and $|\cos(x)| \leq 1$? I can do this easily algebraically/trigonometrically, but I am stuck trying to determine this inequality with series.

Answer 1

To prove that $|\sin(x)|\le 1$ and $|\cos(x)|\le 1$ from the series definition, I would take a little "side step". From $$\sin(x)= \sum_{n= 0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}$$ we can differentiate "term by term" to show that $$(\sin(x))'= (-1)^n\sum_{n= 0}^\infty \frac{2n+1}{2n+ 1}!x^{2n}= \sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{(2n)!}= \cos(x)$$. And from $$\cos(x)= \sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{(2n)!}$$ we get $$(\cos(x))'= \sum_{n= 0}^\infty (-1)^n\frac{2n}{(2n)!}x^{2n-1}= \sum_{n=0}^\infty (-1)^n\frac{1}{2n-1}x^{2n-1}$$. Let $m= n- 1$ so $n= m+ 1$ and that becomes $$\sum_{m=0}^\infty (-1)^{m+1}\frac{1}{(2m+1)!}x^{2m+1}= -\sin(x)$$.

Now, let $$f(x)= \sin^2(x)+ \cos^2(x)$$ then $$f'(x)= 2\sin(x)(\cos(x))+ 2\cos(x)(-\sin(x))= 0$$ for all $x$. Since the derivative is $0$ for all $x$, $f(x)$ is a constant. Taking $x= 0$ we have $$f(0)= 0+ 1= 1$$. Therefore, $$\sin^2(x)+ \cos^2(x)= 1$$ for all $x$.