hey what is the proof of sine and cos series

Assume you can write \(\displaystyle \sin{x}\) as a polynomial.

Then

\(\displaystyle \sin{x} = c_0 + c_1x + c_2x^2 + c_3x^3 + c_4x^4 + \dots\).

Let \(\displaystyle x = 0\) and you can see \(\displaystyle c_0 = 0\).

Differentiate both sides:

\(\displaystyle \cos{x} = c_1 + 2c_2x + 3c_3x^2 + 4c_4x^3 + 5c_5x^4 + \dots\).

Let \(\displaystyle x = 0\) and you can see \(\displaystyle c_1 = 1\).

Differentiate both sides:

\(\displaystyle -\sin{x} = 2c_2 + 3\cdot 2c_3x + 4\cdot 3c_4x^2 + 5\cdot 4c_5x^3 + 6\cdot 5 c_6x^4 + \dots\).

Let \(\displaystyle x = 0\) and you can see \(\displaystyle c_2 = 0\).

Differentiate both sides:

\(\displaystyle -\cos{x} = 3\cdot 2c_3 + 4\cdot 3\cdot 2c_4x + 5\cdot 4\cdot 3c_5x^2 + 6\cdot 5\cdot 4c_6x^3 + 7\cdot 6\cdot 5c_x^4 + \dots\).

Let \(\displaystyle x = 0\) and you can see \(\displaystyle c_3 = -\frac{1}{3\cdot 2} = -\frac{1}{3!}\).

Differentiate both sides:

\(\displaystyle \sin{x} = 4!c_4 + 5!c_5x + 6\cdot 5\cdot 4\cdot 3c_6x^2 + 7\cdot 6\cdot 5\cdot 4c_7x^3 + 8\cdot 7\cdot 6\cdot 5c_8x^4 + \dots\).

Let \(\displaystyle x = 0\) and you can see \(\displaystyle c_4 = 0\).

Differentiate both sides:

\(\displaystyle \cos{x} = 5!c_5 + 6!c_6x + 7\cdot 6\cdot 5\cdot 4 \cdot 3c_7x^2 + 8\cdot 7\cdot 6\cdot 5 \cdot 4c_8x^3 + 9\cdot 8\cdot 7\cdot 6\cdot 5c_9 x^4 + \dots\).

Let \(\displaystyle x = 0\) and you can see \(\displaystyle c_5 = \frac{1}{5!}\).

I think you can see that if we continue this way, we will find

\(\displaystyle \sin{x} = x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \frac{1}{7!}x^7 + \dots - \dots\)

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

A similar process can be used to find the series for \(\displaystyle \cos{x}\). Alternatively, you can differentiate both sides of the series for \(\displaystyle \sin{x}\).