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Homework Help: More Trig Identity Proofs .

  1. Jan 20, 2008 #1
    More Trig Identity Proofs .........

    1. The problem statement, all variables and given/known data
    1. cot^2x - 1 = cot2x
    -----------
    2cotx


    2. tanx + cotx = 2csc2x


    3. cos(A+B) = 1-tanAtanB
    --------- ----------
    cos(A-B) 1+tanAtanB


    2. Relevant equations



    3. The attempt at a solution

    Anyone how to do these proofs??
    i've tried simplifying all of these as much as i could but i don't seem to get the
    right answer ...... can anyone show me???
     
    Last edited: Jan 20, 2008
  2. jcsd
  3. Jan 20, 2008 #2
    What have you tried? Show us your attempts, and we'll guide you toward the answer.

    You might start just by writing the identities that you think might be relevant.
     
  4. Jan 20, 2008 #3
    okay lets work on the first one then ..............

    i'll work on the left hand side.........

    on the top i know that cot^2x = 1 / tan^2x = cos^2x / sin^2x
    and i know that 2cotx is 2 * cosx / sinx

    that is all i have so far ....... what next?? use complex fraction ?? or find common denominator??
     
  5. Jan 20, 2008 #4
    Okay, you need some common identities. cos^2x and sin^2x have a simple relation so that you never need to have both of them - in other words you can always get rid of one and be left with only the other.

    Also, here's a key to problems of this kind: you see that the expressions on one side have functions of x whereas the expressions on the other side have functions of 2x? How you you relate those? You're going to have to be able to get from functions of 2x to functions of x - what identities tie those together?
     
  6. Jan 20, 2008 #5
    ahhhhhhh but there's so many differernt expresions ..... which one should i use ???
     
  7. Jan 20, 2008 #6
    i see that cot2x has the function of 2x .......... so that would make is cos2x / sin2x ..
    then cos2x = cos^2x - sin^2x
    and sin2x = 2sinxcosx so it's now
    cos^2x - sin^2x
    -----------------
    2sinxcosx
     
  8. Jan 20, 2008 #7
    now what about the left side??
    cot^2x - 1
    -----------
    2cotx
     
  9. Jan 20, 2008 #8
    OOOOOOOOO NVM I GOT IT !!!!!!!!!!!!!!!
    how about number 2??

    2. tanx + cotx = 2csc2x
     
  10. Jan 20, 2008 #9
    Those are the ones! Almost everything can be expressed in terms of sines and cosines, so those two
    are your best friends for the rest of the evening.
     
  11. Jan 20, 2008 #10
    Okay, I'll tell you - I have no idea off the top of my head, but I'll bet $5 that if you express everything in terms of sines and cosines you'll be half the way there ....
     
  12. Jan 20, 2008 #11
    got it .................... in sines and cosines
    left side
    ( sinx / cosx ) + ( cosx / sinx )

    right side

    ( 2 / sin2x )

    that's as far as i got ....... but what should i do now ???
    can i find a common denominator for the left side???
    and also can i simplify the right side even more??
     
  13. Jan 20, 2008 #12

    rock.freak667

    User Avatar
    Homework Helper

    When proving Trig Identities, work with one side only. Bring the left side to the same denominator. Then apply some more trig identities that you know.
     
  14. Jan 20, 2008 #13
    Keep going - you've got it. Just simplify- I'd start with the left side and then see if you can get the right side to look the same.
     
  15. Jan 20, 2008 #14
    That's true - the procedure is generally to rearrange one side so that it becomes the same as the other. Sometimes, though, it helps to work "backwards" from the other side a little, and then when you've got the same thing on both sides, to reverse those steps so that you do all the steps on one side.
     
  16. Jan 20, 2008 #15
    okay on the left side I now have
    sin^2x + cos^2x ....
    it can either be
    1 .... which is clearly wrong ....
    or 1-cos^2x + 1 - sin^2x
    Can i re-write this into something else??

    or can it be something else which i have no idea...
     
    Last edited: Jan 20, 2008
  17. Jan 20, 2008 #16
    Okay, I'm not sure how you've gotten where you are. You had:
    ( sinx / cosx ) + ( cosx / sinx )

    What do you get when you add these two? It should be a pretty simple expression.

    Now think about that right hand side - how can you get to that expression? (I'd work it backwards and then reverse your steps to complete the calculation on the left hand side.)
     
  18. Jan 20, 2008 #17
    OOO when i add i get
    ( sinx + cosx / sinx*cosx ) am i right????
     
  19. Jan 20, 2008 #18
    These are the main ones you need to know ...

    http://alt2.mathlinks.ro/Forum/latexrender/pictures/e/0/f/e0f3f955ccf55a86c49f7fa5d4e0454061f7cbdc.gif [Broken]
     
    Last edited by a moderator: May 3, 2017
  20. Jan 20, 2008 #19
    Nope!! Be careful - you had to multiply numerators and denominators by sinx (for the first one) or cosx, right?
     
  21. Jan 20, 2008 #20
    i thought i had to find a common denominator .............. and that would be sinxcosx...
     
  22. Jan 20, 2008 #21
    yargh can u just show me the solution already ...PLEASE!!!... this is really stressing me out waiting for your reply ..... and i have lots of other homework i have to attend to ......
    it would be greatly appreciated
     
  23. Jan 20, 2008 #22
    Keep working at it. Once you get through these concepts your hw will be a breeze.
     
  24. Jan 20, 2008 #23
    Sorry ... I'm bouncing between different threads.

    You did get a common denominator, but you forgot to multiply the numerators by the sinx and cosx factors that you needed to get the denominators the same. Doing that would give you squares of sinx and cosx on top.

    Now go to the RHS and use the identity for sin(a+b) to expand csc(2x) (write it as 1/sin first). Both sides should now look the same.
     
  25. Jan 20, 2008 #24
    alright my left hand side is .....
    (sin^2x*cosx / cosxsinx ) + (sinx*cos^2x / sinxcosx )

    what do u mean by
    "Now go to the RHS and use the identity for sin(a+b) to expand csc(2x) (write it as 1/sin first)"

    my right hand side is
    2 * 1/sin2x = 2/sin2x

    is that what u mean??
     
  26. Jan 20, 2008 #25
    You've got the LHS wrong. You might have to relax and just slow down a bit.

    You should have multiplied sinx/cosx by sinx/sinx, and cosx/sinx by cosx/cosx. That should have given you sin^2x + cos^2x on the top.

    On the RHS you have a sin(2x). What's that equal to? Remember, you have identities for relating sinx to sin(2x)=sin(x+x).
     
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