Hi Lucky,

To develop the derivative of the sine function from first principles requires that yuo show that

\[ \lim_{h \rightarrow 0} \frac{\sin(h)}{h} = 1.\]

I am not going to prove this for you, I suggest that you look in a calculus textbook for a proof.

From first principles

\[f^\prime(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}.\]

For $f(x) = \sin(x)$ this becomes

\[f^\prime(x) = \lim_{h \rightarrow 0} \left( \frac{\sin(x + h) - \sin(x)}{h} \right) .\]

Use of the fact that $\sin(a + b) = \sin(a) \cos(b) + \cos(a) \sin(b)$ and some simplification yields

\[f^\prime(x) = \lim_{h \rightarrow 0} \left( \sin(x) \frac{\cos(h) - 1}{h} + \cos(x) \frac{\sin(h)}{h} \right).\]

Since we know that

\[ \lim_{h \rightarrow 0} \frac{\sin(h)}{h} = 1.\]

it remains to evlauate

\[ \lim_{h \rightarrow 0} \frac{\cos(h) - 1}{h}.\]

Multiplying the numerator and denominator by $\cos(h) + 1$ and using the fact that $\sin^2(h) = 1 - \cos^2(h)$ this becomes

\[ \lim_{h \rightarrow 0} \frac{\cos(h) - 1}{h} = \sin(x) \; \lim_{h \rightarrow 0} \frac{- \sin^2(h)}{h \cos(h) + 1}\]

which simplifies to

\[ \lim_{h \rightarrow 0} \frac{\cos(h) - 1}{h} = -\lim_{h \rightarrow 0} \frac{\sin(h)}{h} ) \times \lim_{h \rightarrow 0}\frac{\sin(h)}{\cos(h) + 1} = -1 \times \left( \frac{0}{1 + 1} \right) = 0.\]

Thus

\[f^\prime(x) = \lim_{h \rightarrow 0} \sin(x) \frac{\cos(h) - 1}{h} + \cos(x) \frac{\sin(h)}{h} = 0 + \cos(x) \times 1 = \cos(x).\]

Harley