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!

Help on trig identity

  1. Apr 14, 2009 #1

    Kiff

    User Avatar

    Help plz on trig identity

    1. The problem statement, all variables and given/known data
    Simplify cot2xsecx + 1/cosx





    3. The attempt at a solution
    Well so far i got:
    cot2xsecx + 1/cosx
    =(cos2x/sin2x)(1/cosx) + 1/cosx
    =((1+cos2x)/(1-cos2x))(1/cosx) + 1/cosx


    and from there im stuck, I've tried playing around with it but I just seem to get to dead ends.
    any help is appreciated, Thanks
     
    Last edited: Apr 14, 2009
    Kiff, Apr 14, 2009
  2. jcsd
    Phys.org - latest science and technology news stories on Phys.org
  3. Apr 14, 2009 #2

    Cyosis

    User Avatar
    Homework Helper

    Re: Help plz on trig identity

    In your second step cancel the square of the cosine. Then merge the entire term into one fraction.
     
    Cyosis, Apr 14, 2009
  4. Apr 14, 2009 #3

    Kiff

    User Avatar

    Re: Help plz on trig identity

    wow that was fast, thanks

    so by cancelling the square I get:
    (cosx/sinx)(1/cosx) + 1/cosx
    =sinx/cos2x + 1/cosx <--- multiply by cosx
    =sinx/cos2x + cosx/cos2x
    =sinx + cosx

    did i do that correctly?
    the question im doing is multiple choice answer tho
    a. csc2xsecx
    b. sec3x
    c. sec2xcscx
    d. csc3x
     
    Kiff, Apr 14, 2009
  5. Apr 14, 2009 #4

    whybother

    User Avatar

    Re: Help plz on trig identity

    [tex] \cot^2{x} \sec{x} + {1 \over \cos{x}} [/tex]

    = [tex] \cot^2{x}\sec{x} + \sec{x} [/tex]

    = [tex](\cot^2{x} + 1) \sec{x} [/tex]

    = [tex]\csc^2 {x} \sec{x} [/tex]

    If you are familiar with the identity, [tex] \cot^2{x} + 1 = \csc^2 {x} [/tex], it's obvious.

    EDIT: In case you aren't familiar with that particular identity, it's simple to show with a couple more steps:
    [tex] 1 + cot^2{u} [/tex]

    =[tex]1 + {cos^2{u} \over sin^2{u}} [/tex]

    =[tex]{sin^2{u} + cos^2{u} \over sin^2{u}} [/tex]

    =[tex]{1 \over \sin^2{u} } [/tex]

    =[tex]\csc^2{u} [/tex]
     
    Last edited: Apr 14, 2009
    whybother, Apr 14, 2009
  6. Apr 14, 2009 #5

    whybother

    User Avatar

    Re: Help plz on trig identity

    Also, in you solution, this step is not valid:

     
    whybother, Apr 14, 2009
  7. Apr 14, 2009 #6

    Kiff

    User Avatar

    Re: Help plz on trig identity

    awsome..thanks guys, omfg i forgot bout that identity
     
    Kiff, Apr 14, 2009
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Help on trig identity

  1. Simplifying trig Ident. Help (Replies: 16)

  2. Trig Ident. Help (Replies: 1)

  3. Help verifying a trig identity? (Replies: 2)

  4. Help with Trig Identity Simplification (Replies: 3)

  5. Help with trig identities (Replies: 3)

Loading...