Proof: The Derivative of sin(x) in Radians

Using the formula for differentiation by first principles:

\[\frac{d}{dx} \sin x = \lim_{h \to 0} \frac{\sin(x+h)-\sin x}{h}\]

We can expand \(\sin(x+h)\) using the addition formulae, giving:

\[\frac{d}{dx} \sin x = \lim_{h \to 0} \frac{\sin x \cos h + \sin h \cos x -\sin x}{h}\]

We know that \(sin x \approx x\) and \(\cos x \approx 1-\frac{x^2}{2}\) (proofs of these can be found here). Therefore we can use these to substitute for \(\sin h\) and \(\cos h\) like so:

\[\frac{d}{dx} \sin x = \lim_{h \to 0} \frac{\left(1-\frac{h^2}{2}\right)\sin x + h \cos x -\sin x}{h}\]

\[\frac{d}{dx} \sin x = \lim_{h \to 0} \frac{\sin x-\left(\frac{h^2}{2}\right)\sin x + h \cos x -\sin x}{h}\]

\[\frac{d}{dx} \sin x = \lim_{h \to 0} \frac{-\left(\frac{h^2}{2}\right)\sin x + h \cos x}{h}\]

\[\frac{d}{dx} \sin x = \lim_{h \to 0} \left( -\left(\frac{h}{2}\right)\sin x + \cos x \right)\]

\[\frac{d}{dx} \sin x = \cos x\]

\[Q.E.D\]

 

A similar argument can be used to show that \(\frac{d}{dx} (\cos x)=-\sin x\) .

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