Hi Saskia,

I assume that $x$ is in degrees also. The differentiation fact that we know about the differentiation of the sine function is that

\[\frac{d}{dx} \left(\sin(x)\right) =\cos(x)\]

is only valid if x is in radians. To differentiate a trig function where the variable is given in degrees we first need to convert the degree measure to radians. I remember the conversion as $180^o = \pi \mbox{ radians}$ and hence $1^o = \large \frac{\pi}{180}$ radians.

Thus if $z$ is in degrees and $f(z) = \sin(z)$ then converting to radians

\[f(z) = \sin\left(\frac{\pi}{180} z\right)\]

and then

\[f^\prime(z) = \left[\cos\left(\frac{\pi}{180} z\right)\right]\times \frac{\pi}{180}.\]

Simplifying and going back to the trig functions that operates in degrees I get

\[f^{\; \prime}(z) = \frac{\pi}{180} \cos(z)\]

I hope this helps,
Harley