1. Not finding help here? Sign up for a free 30min 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!

Trig Identity

  1. Dec 2, 2009 #1
    1. The problem statement, all variables and given/known data

    verify the following identity:

    Sec(x)Sin2(x)
    ______________________ = 1 - cos(x)

    1 + sec(x)


    2. Relevant equations
    sec(x)=1/cos(x)
    sin2(x)=1-cos2(x)



    3. The attempt at a solution
    I never know how to start off these problems. I have to take the left side and show that it equals the right by doing trig identities. I try several steps and keep going through an endless loop. I believe I must start of by multiplying 1-sec(x)/1-sec(x)
    That way i get

    (1-sec(x))(sec(x)sin2(x))
    __________________________________

    1 - sec2(x)

    I'm not sure if i'm starting off correctly

    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 2, 2009 #2

    Mark44

    Staff: Mentor

    It would probably be easier to start by replacing the sec(x) terms on the left side with 1/cos(x).
     
  4. Dec 2, 2009 #3
    Sec(x)Sin2(x)
    ______________________ = 1 - cos(x)

    1 + sec(x)

    ok using the identity sec(x)=1/cos(x)

    I get

    1/cos(x) * sin2(x)
    __________________________

    1 + 1/cos(x)

    and that =

    sin2(x)/cos(x)
    ______________________

    1 + 1/cos(x)

    multiply top and bottom by cos(x)

    Sin2(x) / 2

    Did i do anything wrong?


    Thanks
     
  5. Dec 2, 2009 #4

    Mark44

    Staff: Mentor

    Yes. cos(x)*(1 + 1/cos(x)) is not equal to 2. Try again.
     
  6. Dec 2, 2009 #5
    I think a trick in handling trigo qns is to convert all to sine and cosine functions.
    since sec, csc and even tan functions can be derived from sine and cosine.

    Do remember this to help in your future sch works as well.
     
  7. Dec 2, 2009 #6
    attachment.php?attachmentid=22239&stc=1&d=1259808742.jpg
     

    Attached Files:

  8. Dec 2, 2009 #7
    Turn the denominator into [tex]\frac{cos(x)+1}{cos(x)}[/tex].

    So that would look like [tex]\frac{\frac{sin^{2}(x)}{cos(x)}}{\frac{cos(x)+1}{cos(x)}}.[/tex]

    What icystrike showed is the faster way but since you're already this far, you can try what I suggested.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Trig Identity
  1. Trig identities (Replies: 2)

  2. Trig Identity (Replies: 7)

  3. Trig identities (Replies: 10)

  4. Trig Identities (Replies: 8)

  5. Trig Identities (Replies: 18)

Loading...