Sin Derivative Proof

[Jason76]

Post Edited

The professor was working on this today in class.

\(\displaystyle f(x) = \sin(x)\)

\(\displaystyle f'(x) = \dfrac{d}{dx} \sin(x) = \cos(x)\)

Proof

Using: \(\displaystyle \lim h \to 0[\dfrac{(x + h) - (f(x))}{h}]\)

\(\displaystyle \lim h \to 0[\dfrac{\sin(x + h) - (\sin(x))}{h}]\)

\(\displaystyle \lim h \to 0[\dfrac{[\sin(x)\cos(h) + \cos(x)\sin(h)] - (\sin(x))}{h}]\) using sum and difference identity \(\displaystyle \sin(A + B) = \sin(A) \cos(B) + \cos(A) \sin(B)\) for \(\displaystyle \sin(x + h)\) Next move

 

[srmichael]

The professor was working on this today in class.

\(\displaystyle f(x) = \sin(x)\)

\(\displaystyle f'(x) = \dfrac{d}{dx} \sin(x) = \cos(x)\)

Proof

Using: \(\displaystyle \lim h \to 0[\dfrac{(x + h) - (f(x))}{h}]\)

\(\displaystyle \lim h \to 0[\dfrac{\sin(x + h) - (\sin(x))}{h}]\)

\(\displaystyle \lim h \to 0[\dfrac{\sin(x)\cos(x) + \cos(x)\sin(x) - (\sin(x))}{h}]\) using sum and difference identity \(\displaystyle \sin(A + B) = \sin(A) \cos(B) + \cos(A) \sin(B)\) for \(\displaystyle \sin(x + h)\) ====> WRONG! (See below)Next move

sin(x+h) = sin(x)cos(h) + cos(x)sin(h)

 

[Jason76]

sin(x+h) = sin(x)cos(h) + cos(x)sin(h)

Original post edited

 

[srmichael]

\(\displaystyle lim_{h \rightarrow 0}\dfrac{sin(x)cos(h)+cos(x)sin(h)-sin(x)}{h}\)

\(\displaystyle =lim_{h \rightarrow 0}\dfrac{sin(x)cos(h)-sin(x)+cos(x)sin(h)}{h}\)

\(\displaystyle =lim_{h \rightarrow 0}\dfrac{sin(x)[cos(h)-1]+cos(x)sin(h)}{h}\)

\(\displaystyle =lim_{h \rightarrow 0}\left(\dfrac{sin(x)[cos(h)-1]}{h}+\dfrac{cos(x)sin(h)}{h}\right)\)

Can you take it from here? Hint: Use the two limit rules for trig that you learned, namely:

\(\displaystyle lim_{h \rightarrow 0}\dfrac{sin(h)}{h}=1 \ \ and\ \ lim_{h \rightarrow 0}\dfrac{1-cos(h)}{h}=0\)

 

[pka]

Original post edited

\(\displaystyle \\\dfrac{[\sin(x)\cos(h) + \cos(x)\sin(h)] - (\sin(x))}{h}=\\\sin (x)\left( {\dfrac{{\cos (h) - 1}}{h}} \right) + \cos (x)\left( {\dfrac{{\sin (h)}}{h}} \right)\)

 

[Jason76]

\(\displaystyle \\\dfrac{[\sin(x)\cos(h) + \cos(x)\sin(h)] - (\sin(x))}{h}=\\\sin (x)\left( {\dfrac{{\cos (h) - 1}}{h}} \right) + \cos (x)\left( {\dfrac{{\sin (h)}}{h}} \right)\)

What are u doing here exactly?

 

[Deleted member 4993]

Algebra.......

Sir Michael Wrote

\(\displaystyle lim_{h \rightarrow 0}\dfrac{sin(x)cos(h)+cos(x)sin(h)-sin(x)}{h}\)

\(\displaystyle =lim_{h \rightarrow 0}\dfrac{sin(x)cos(h)-sin(x)+cos(x)sin(h)}{h}\)

\(\displaystyle =lim_{h \rightarrow 0}\dfrac{sin(x)[cos(h)-1]+cos(x)sin(h)}{h}\)

\(\displaystyle =lim_{h \rightarrow 0}\left(\dfrac{sin(x)[cos(h)-1]}{h}+\dfrac{cos(x)sin(h)}{h}\right)\)

Can you take it from here? Hint: Use the two limit rules for trig that you learned, namely:

\(\displaystyle lim_{h \rightarrow 0}\dfrac{sin(h)}{h}=1 \ \ and\ \ lim_{h \rightarrow 0}\dfrac{1-cos(h)}{h}=0\)

Read it carefully.....

 

[Jason76]

Ok, I see what's going on, will finish soon.

 

[HallsofIvy]

Essentially, this proof is using the facts that \(\displaystyle \lim_{x\to 0}\frac{sin(x)}{x}= 1\) and \(\displaystyle \lim_{x\to 0}\frac{1- cos(x)}{x}= 0\) which are typically proved with a rather hand-waving "geometric" proof.

It is also possible to define \(\displaystyle sin(x)= \sum_{n= 0}^\infty \frac{(-1)^n}{(2n+1)!} x^{2n+1}\) and \(\displaystyle cos(xZ)= \sum_{n= 0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}\), showing that those sums converge uniformly over any finite integral, and determine the derivatives by differentiating "term by term".