A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

Hi, In lecture 7 it is stated that, Sqaureroot [ ( 1 - cost)^2 + sin^2t ] is equal to Sqaureroot [ 1- 2cost + cos^2t + sin^2t ] I don't understand where the 2cost in the second expression came from. I've tried playing around with some trig identifies with no luck. Could someone please explain it? Thanks.

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it's simple try (a-b)^2=a^2+b^2-2ab

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    That's great. Thanks keerthy.

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

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.