Question

Part A, Session 6: Calculating Derivatives
At 34:04 the professor is claiming that the gap is so tiny. This is not mathematical prove, is it? It looks smaller, but don't we need to proof mathematically that the gap (1-cos(theta) ) is smaller than length theta?
I have asked this question in the following link, but I did not get any answer.
http://openstudy.com/updates/4f710727e4b0eb8587738591

Answer 1

It is just a visual representation of the short angle theory and why it is that we can claim that the vertical line is the same length as the arc.

Answer 2

I agree with sin proof but I am not convinced how short angle theory is applied to cosine.

Answer 3

the cos proof is actually derived from what we know about the sine limit
ie
\[\lim_{\theta \rightarrow 0}(\cos \theta-1)/\theta\]
can be multiplied by \[(\cos \theta+1)/(\cos \theta+1)\]
giving \[\lim_{\theta \rightarrow 0}(\cos^{2}-1 )/\theta(\cos \theta+1)\]
\[ \lim_{\theta \rightarrow0}(-\sin^{2} \theta)/\theta(\cos \theta+1)\]
put away the constant -1
\[ -\lim_{\theta \rightarrow0}(\sin^{2} \theta)/\theta(\cos \theta+1)\]
we can re-write the fraction as
\[-\lim_{\theta \rightarrow 0}\left(\sin \theta/\theta \right)*\sin \theta/(\cos \theta+1) )\]
\[-\lim_{\theta \rightarrow 0}\left(\sin \theta/\theta \right)*\lim_{\theta \rightarrow 0}(\sin \theta/(\cos \theta+1) )\]
since\[\lim_{\theta \rightarrow 0}\left(\sin \theta/\theta \right)=1\]
\[\lim_{\theta \rightarrow 0}(\sin \theta)=0\]
and
\[\lim_{\theta \rightarrow 0}(\cos \theta+1) =1+1\]
we can substitute everything to give
\[(-1)*(0/1+1)=0\]
\[\lim_{\theta \rightarrow 0} (\cos \theta-1) /\theta=0\]

Answer 4

Very nice. :)