$\cos(47^\circ)\sin(32^\circ)$ approximation by differentials I need to approximate $\cos(47^\circ)\sin(32^\circ)$. In order to do this, I need to use differentials. So, for a function $f(x,y)$, we have:
$$f(x,y)-f(x_0,y_0)\approx \frac{\partial f(x_0,y_0)}{\partial x}(x-x_0) + \frac{\partial f(x_0,y_0)}{\partial y}(y-y_0)$$
When we let $(x,y)$ get close to $(x_0,y_0)$ we can say that:
$$\delta f \approx \frac{\partial f(x_0,y_0)}{\partial x}\delta x + \frac{\partial f(x_0,y_0)}{\partial y}\delta y$$
Then we define the differential as being
$$df = \frac{\partial f(x_0,y_0)}{\partial x} dx +\frac{\partial f(x_0,y_0)}{\partial y}dy $$
In the exercise, I choose:
$$(x,y) = (47^\circ, 32^\circ), (x_0, y_0) = (45^\circ, 30^\circ)$$
So we should have:
$$\delta f \approx \frac{\partial f(45^\circ,30^\circ)}{\partial x}(47^\circ-45^\circ) + \frac{\partial f(45^\circ,30^\circ)}{\partial y}(32^\circ-30^\circ)$$
but this $(47^\circ-45^\circ)$ and $(32^\circ-30^\circ)$ doesn't seem rigth, because we're working with degress, and $\delta z$ is in radians. I'm quite confused in how to solve this exercise. Could somebody give show me what's happening?
 A: Since $x,y$ are in degrees, the chain rule gives:
$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial x} \frac{\pi}{180}= -sin(x)sin(y)\frac{\pi}{180}$
$\frac{\partial f}{\partial y} = \frac{\partial f}{\partial y} \frac{\pi}{180}= cos(x)cos(y)\frac{\pi}{180}$
Therefore, the linear approximation for $f$ is:
$f(x,y) \approx f_0 + \frac{\pi}{180}\left.\frac{\partial f}{\partial x}\right|_0 (x-x_0) + \frac{\pi}{180}\left.\frac{\partial f}{\partial y}\right|_0 (y-y_0)$
or 
$f(x,y) \approx cos(x_o)sin(y_o) - \frac{\pi}{180}sin(x_o)sin(y_o)(x-x_0) + \frac{\pi}{180}cos(x_o)cos(y_o)(y-y_0)$
In your case:
$f(47^{\circ},32^{\circ}) \approx cos(45^{\circ})sin(30^{\circ}) - \frac{\pi}{180}sin(45^{\circ})sin(30^{\circ})(47^{\circ}-45^{\circ}) + \frac{\pi}{180}cos(45^{\circ})cos(30^{\circ})(32^{\circ}-30^{\circ}) $
$ =\frac{\sqrt{2}}{2}\cdot\frac{1}{2} + \frac{\pi}{180}\left(-\frac{\sqrt{2}}{4}\cdot2 + \frac{\sqrt{6}}{4}\cdot 2\right)  = 0.3626^c = 20.77^\circ$
A calculator gives $f(47^{\circ},32^{\circ}) = cos(47^{\circ})sin(32^{\circ}) = 0.3614^c = 20.71^\circ$, indicating an error of order $0.1^\circ $ in the linear approximation.
A: You can convert degrees to radians by  multiplying by $\frac \pi {180}$ so $2^\circ=\frac {2\pi}{180}\approx 0.0345$ radans
