# Basic calculus 1 question: What is the defining formula of sinΘ?

## Homework Statement

Q) Derive the dereitvative of sinΘ using the definition of differentiation.

I am in Calculus 3, and I used to know how to work this problem very well! :) I just don't remember lol - I just need a little help I guess! :)

## Homework Equations

lim [f(x+h) - f(x))]/h
x→0

## The Attempt at a Solution

f(x) = sinΘ

f(x+h) = (sinΘ + h)

Now I am STUCK!! lol

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rock.freak667
Homework Helper
well now put that into the formula and check the numerator.

You will have sin(θ+h)-sinθ. Do you know your sum to product formulas? These will help greatly here or you can expand sin(θ+h) as well.

Mark44
Mentor

## Homework Statement

Q) Derive the dereitvative of sinΘ using the definition of differentiation.

I am in Calculus 3, and I used to know how to work this problem very well! :) I just don't remember lol - I just need a little help I guess! :)

## Homework Equations

lim [f(x+h) - f(x))]/h
x→0

## The Attempt at a Solution

f(x) = sinΘ

f(x+h) = (sinΘ + h)

Now I am STUCK!! lol

Might as well use x instead of theta, since x is easier to type.

f(x + h) = sin(x + h)
Use the identity for the sine of the sum of two angles.

You will also need a limit:
$$\lim_{h \to 0} \frac{sin(h)}{h}$$
Do you remember the value of this limit?

so, I write:

lim [sin(x+h) - sin(x))]/h
x→0

So, I get:

lim [sin(x) cos(h) + cos(x)sin(h) - sin(x))]/h
x→0

But now I am stuck again lol - I understand that if I use $$\lim_{h \to 0} \frac{sin(h)}{h}$$

then my equation should become:

lim [sin(x) cos(h) + cos(x)sin(h) ]/h , right?
x→0

Mark44
Mentor
Your limit should be as h --> 0, not as x --> 0. Also, you are missing a term in the numerator. What happened to the -sin(x) that used to be there?

There's another limit that you'll need as well:
$$\lim_{h \to 0} \frac{cos(h) - 1}{h}$$

Took me a while for me to figure this out!! I spent time expanding on the ideas both of you gave me and finally understood what to do!!!

THANKS to BOTH of you!!!!!!!!!! :)