Deriving power series for $\sin x$ without using Taylor's Theorem or $\exp z$ Starting with defining $(\cos t, \sin t)$ from the unit circle, is it possible to derive the power series for $\sin(t)$: $$\sin t = t

Deriving power series for $\sin x$ without using Taylor's Theorem or $\exp z$ Starting with defining $(\cos t, \sin t)$ from the unit circle, is it possible to derive the power series for $\sin(t)$: $$\sin t = t - \frac{t^3}{3!} + \frac{t^5}{5!} - \dots$$ Note: I will be answering my question, I hope this doesn't offend anyone. If is any issue with the proof, I am grateful for any improvement.

Another method is using the inequality

$$\sin(x)\leq x$$

Integrate this between $0$ and $t$

$$-\cos(t)+\cos(0)\leq t^2/2$$ $$1-t^2/2\leq \cos(t)$$

Integrate this between $0$ and $x$:

$$x-x^3/3!\leq \sin(x)$$

Repeat this and you will get a lower and upper bound, which are just partial sums of the Taylor series. Note that you directly get the taylor series for $\cos(x)$.