1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Trig Identities

  1. Nov 18, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove the following identity:

    cos2A/1 + sin2A = cotA - 1 / cotA + 1

    2. Relevant equations

    3. The attempt at a solution

    I proved the right side, which eventually lead up to cosA - sinA / cosA + sinA

    I have NO idea how to do the left side. I have wasted roughly 10-15 sheets of paper now trying to prove it, but it all leads up to nothing. I tried MANY different solutions...
  2. jcsd
  3. Nov 18, 2008 #2
    did u use the double angle formulas of sin2u=2sinucosu and cos2u=1-2sin^2u?
  4. Nov 18, 2008 #3
    http://img141.imageshack.us/img141/8388/trigom8.jpg [Broken]
    Last edited by a moderator: May 3, 2017
  5. Nov 18, 2008 #4
    Vitaly, thanks for trying BUT I have a question about where sin^2a(cot^2 - 1) came from.. could you add me to MSN please? supaflygt@hotmail.com
  6. Nov 18, 2008 #5
    http://img139.imageshack.us/img139/9676/trigkw2.jpg [Broken]
    no probs :wink:
    Look carefully, it's sine squared, not sin(2A). My 2A are big
    Last edited by a moderator: May 3, 2017
  7. Nov 18, 2008 #6
    Yeah, I know. I'm curious how you got sin^2a/sin^2a + sin^2a in the equation (2nd step) :P
  8. Nov 18, 2008 #7
    Please use brackets where necessary. (sinA)^2/(sinA)^2=1 and as you must know if you multiply something by 1, it doesn't change anything. It's a very common trick in math, e.g. integrating logs, taking inverse of a matrix, etc..
    So, I muultipled (cosA)^2 by 1. And (sinA)^2 is therefore a common factor.
    It's like (5+7)=(5*7/7+7)=7(5/7+1)
  9. Nov 18, 2008 #8
    Well, I tried it doing my own thing, and here is the result:
    EDIT: it's not -1 or 0, but it must be something.
  10. Nov 18, 2008 #9
    Solved it now, thanks for your help.
  11. Nov 19, 2008 #10
    my friend they are not equal to negative one. however you can prove both sides are
    (cos A-sin A)/(cos A+sin A)
  12. Nov 19, 2008 #11
    My God, seeing those divisions just scares me.
    Revisit fractions please and formulas of (a+b)2, a2-b2

    And consider this: (52-32)/(5+3)2 is NOT equal -1.
    (52-32)/(5+3)2=(5-3)(5+3)/((5+3)(5+3)). Divide by a common factor of 5+3, leaving you with (5-3)/(5+3)

    Learn (a+b)2=(a+b)(a+b)=a2+2ab+b2, a2-b2=(a-b)(a+b) NOW
  13. Nov 19, 2008 #12
    I know they aren't equal to -1. I proved both sides were (cosa-sina)/(cosa+sina), though, which was why I said it was solved
  14. Nov 20, 2008 #13
    He says so because of your amazing cancellation.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook