



 
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,  


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