In the differentiation of trigonometric function which is not in degrees,why are we using a conversion factor.Will the formulas only work for radian measures?

For example: x is in degrees,what is d(sin x)/dx?
So, the derivative is (pi/180)*cos(x). Why is it so?

Question

In the differentiation of trigonometric function which is not in degrees,why are we using a conversion factor.Will the formulas only work for radian measures?  

For example: x is in degrees,what is d(sin x)/dx?
So, the derivative is  (pi/180)*cos(x). Why is it so?

anonymous · Answer

I thought that radians only mattered in the proof. Theta was being used as both the angle and the length of the arc. The cos of 0 degrees is 1, which corresponds to the d(sin 0). The slope seems right at 90 and 180 degrees. I'm only up to the middle of lecture 4, so maybe I missed something.

anonymous · Answer

No!!! All the calculus works only for the radians.. No way for degrees.. Remember using, 

arc length = theta * radius ... somewhere while deriving for the derivative of sin (theta)... That wouldn't hold for degree, isn't it?

anonymous · Answer

I can see why it makes a difference in the proof. Once you have the derivative though, why would it matter? If I take D sin x and get cos x, switching my calculator from degrees to radians is going to give me the same answer to 90 degrees or pi/2 respectively.

anonymous · Answer

Well ariyama check again.It will only work if you change the mode of the calculator from degrees to radians.So when you do the changing the calculator automatically adds a conversion factor.

anonymous · Answer

kartiksrimak, Thanks, that was my point. Once the derivative is taken, the units don't matter as long as you use them consistently. Using radians does matter when constructing the proof. That is all I was attempting to say. Sorry to belabor the point.

anonymous · Answer

please for the love of god be careful if you are not 100% sure

in calculus you ALWAYS use radians