• anonymous
I'm having trouble visualizing why in the geometric proof of the derivative of sine the two angles are congruent. Professor Jerison does something about rotating by 90 degrees the triangle but i quite don't understand why. This proof is given in lecture 3.
OCW Scholar - Single Variable Calculus
  • schrodinger
See more answers at
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly


Get your free account and access expert answers to this
and thousands of other questions

  • phi
Yes, I find that "explanation" unsatisfying we could make this argument:|dw:1436789799775:dw| where I used "a" rather than theta. and not a (a with bar) to represent 90-a
  • phi
now form the triangle we are interested in |dw:1436790048402:dw|
  • phi
btw, here is a more clear proof by Khan

Looking for something else?

Not the answer you are looking for? Search for more explanations.