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 Question from Saskia, a student: derivative of y = sin (30º + x)

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

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