Deriving the Derivative of the sine of an angle.

To the right is the definition of a limit. We will use it to show that the derivative of the sine of angle \(x\) is the cosine of angle \(x\).

$$\lim_{h \to 0} {f(x + h) - f(x) \over h}$$

Let \(f(x) = sinx\), thus \(f(x + h) = sin(x + h)\). By trigonometric identity, \(sin(x + h) = cosxsinh + sinxcosh\). Substituting these quantities into the general formula we get:

$$\lim_{h \to 0} {cosxsinh + sinxcosh - sinx \over h}$$

By addition rule of limits, we get:

$$\lim_{h \to 0} {cosxsinh \over h} + \lim_{h \to 0} {sinxcosh - sinx \over h}$$

Factoring out \(sinx\) from \(sinxcosh - sinx \over h\), we get:

$$\lim_{h \to 0} {cosxsinh \over h} + \lim_{h \to 0} {sinx(cosh - 1) \over h}$$

As \(h \to 0\), two things are occurring in the previous formula, \(sinh \over h\) \(\approx 1\) and \(cosh \approx 1\), thus making \(1 - cosh = 0 \), so we get:

$$\lim_{h \to 0} {cosx} + \lim_{h \to 0} {0}$$

Since \(cosx\) and \(0\) are constants, we can remove the limits and we finally get:

$${d \over dx} {sinx = cosx}$$