Differentiating trig functions

The derivative of a function of a real variablemeasures the sensitivity to change of a quantity (a function value or dependent variable) which is determined by another quantity (the independent variable). Derivatives are a fundamental tool ofcalculus. For example, the derivative of the position of a moving object with respect to time is the object's velocity: this measures how quickly the position of the object changes when time is advanced.The derivative of a function of a single variable at a chosen input value is the slopeof the tangent line to the graph of the functionat that point. This means that it describes the best linear approximation of the function near that input value. For this reason, the derivative is often described as the "instantaneous rate of change", the ratio of the instantaneous change in the dependent variable to that of the independent variable.Hope this helps

Why does X have to be in radians when differentiating or integrating trig functions?


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hilarious. not.

Too advanced for you?

Why does X have to be in radians when differentiating or integrating trig functions?


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Too advanced for you?

because 'sine' etc. as a function on the real numbers uses radians.

This isn't entirely true I'm afraid!

It's perfectly possible to define sine etc as a function on the real numbers using degrees. so that

0 -> 0
30 -> 1/2
90 -> 1

etc

The real reason for using radians in differentiation is that somewhere along the line you end up relying on the limit of (sin x) / x being 1 as x->0 and this is only true if x is measured in radians.

(Of course this is the 'A level justification'. At university sine etc are defined by power series, so the issue becomes irrelevant!)

I don't understand why what you say is only true when X is measured in radians? Surely sin(0.08 degrees)/0.08degrees is basically 1?


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Not according to my calculator.

sin(0.08 degrees)/0.08 is approximately 0.01745

If x is measured in degrees, then as x gets closer to 0, the ratio (sin x)/x gets closer to pi/180 which is approximately 0.017453. That's why you need to use radians for differentiation - otherwise you'd have loads of annoying (pi/180) factors to include!

But isn't sin(nearly 0) = 0? So it wouldn't it be nearly0/180 =0 not =1


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Oh I think I get it. Like if you use rads you can just say pi/pi = 1


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I don't understand why what you say is only true when X is measured in radians? Surely sin(0.08 degrees)/0.08degrees is basically 1?


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But isn't sin(nearly 0) = 0? So it wouldn't it be nearly0/180 =0 not =1


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So to say sin(X-180) differentiated = cos(X-180) would be incorrect but to say it was cos(X-pi) would be correct? How are these answers any different?


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No, we're talking about a limit. As x gets closer to 0, so does sin x. So on the face of it, as x gets close to 0 it looks like you're calculating 0 / 0. However, we can show (and it should be in any decent A level textbook!) that if x is measured in radians, then (sin x / x) approaches 1 as x gets closer and closer to 0.




I can't make much sense of what you're asking there - you can't mix up radians and degrees! If x is measured in radians then the derivative of sin x is cos x. Outside GCSE work (and practical things like surveying, navigation etc) no-one uses degrees with trigonometry.

If your 180 is a quantity in degrees, then sin(x - 180) only makes sense if x is in degrees too, so its derivative is certainly not cos(x - 180)!

You really need to find a decent A level textbook that covers differentiation of trig functions from first principles because it covers things like the limit of (sin x)/ x using a diagram which squeezes this quantity between 2 other limits that are easier to evaluate.

I just don't get why it matters whether degrees or radian are used because if you differentiate eg sin(X-pi) then you get cos(X-pi). Couldn't I choose to write this as cos(180x/pi -180) and that would mean the same thing?


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Only if x is in radians.

Do you understand how the derivative is defined in terms of a limit?

Doesn't your A level textbook show you how the derivative of sin x is arrived at from first principles?

Yes I do understand that. It's based on the fact that as X approach 0, sinx approaches X. So then you can cancel these out. I just don't see why this wouldn't also be true for degrees.


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