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Trig identity questions

  1. Jun 10, 2013 #1
    Hey guys,I need some help on the following trig identities:

    1) sin2x = 2tanx/1+tan^2x

    2) sin2x/sinx - cos2x/cosx = secx

    My attempts:

    1) LS: sin2x
    2sinxcosx

    2sinx/cosx

    2tanx/1+tanx

    Not sure if this is right or not. I kind of understand my third step but it just doesn't seem right, any validations? If not, help please.

    2) LS: sin2x/sinx - cos2x/cosx

    2sinxcosx/sinx - ???

    sinx/cosx - ?? i have no idea where to go from here
     
  2. jcsd
  3. Jun 10, 2013 #2

    verty

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    Do #1 in reverse, it should be easier, and take little steps, don't make your steps too large. Each step should be small enough to be obvious to the reader.
     
  4. Jun 10, 2013 #3

    HallsofIvy

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    Yes, sin(2x)= 2sin(x)cos(x)

    What? 2 sin(x)/cos(x)= 2 tan(x). Have you changed to the right side now??


    Where did this come from? It is what you want to arrive at but what are you doing to get it?

    Frankly, I don't understand what you are trying to do. Please state exactly what you are doing to go from one step to another.
     
  5. Jun 10, 2013 #4
    I just stated that I was trying to turn the left side into the right, sorry for the confusion. When I do these I just pick a side then use identities until it is equal to the opposite side.
     
  6. Jun 10, 2013 #5

    tiny-tim

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    hi tg22542! :smile:

    (try using the X2 button just above the Reply box :wink:)
    as verty :smile: says, it would be a lot easier to go from right to left!

    if you must go from left to right, use cos = 1/sec :wink:
    first step is obviously to put them over a common denominator
     
  7. Jun 10, 2013 #6
    okay so I believe I completed 1)

    RS:

    2tanx/1+tan^2x

    2tanx/sec^2x

    2sinx2cosx/sec^2x

    2sinxcosx

    sin2x

    good? :)

    thank you guys very much for your help so far, very appreciated
     
  8. Jun 10, 2013 #7
    nevermind.. just realized the identity I used for sec is actually sec^2..

    ugh
     
  9. Jun 10, 2013 #8

    MarneMath

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    Looks good to me.

    For your other question, do you know any trig identities dealing with cos[2x]? Such as Cos[2x] = Cos[x]^2 - Sin[x]^2 ?
     
  10. Jun 10, 2013 #9

    Mark44

    Staff: Mentor

    The expressions below are all equal, so you should denote them as such by using =.

    Also, what you wrote below needs parentheses around the terms in the denominator. What you wrote is really this:
    $$\frac{2tan(x)}{1} + sec^2(x)$$
     
  11. Jun 11, 2013 #10

    tiny-tim

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    yes :smile:
    where did this come from? :confused:

    use 1/sec = cos (as i said before)​
     
  12. Jun 11, 2013 #11
    So could it go :

    RS: 2tanx/1+tan^2x

    =sin^2x/cos^2x + 1/cos^2x ??
     
  13. Jun 11, 2013 #12

    tiny-tim

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    (try using the X2 button just above the Reply box :wink:)
    i honestly have no idea how you got that :confused:

    tg22542, in the exam, you must do one step at a time …

    then the examiner can see what you're doing, and why, and at least give you some marks for getting something right!

    try again, slowwwwly :smile:
     
  14. Jun 11, 2013 #13
    RS: 2tanx/1+tan^2x

    So I first used the identity tan^2x + 1 = sec^2x

    So

    = 2tanx + sec^2x

    Next I used tanx=sinx/cosx

    So:

    2sinx/2cosx + sec^2x

    Now I'm lost, It must equal 2sinx so I clearly have to get the cos to cancel out somehow, but I don't know which identities to use
     
  15. Jun 11, 2013 #14

    tiny-tim

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    correct :smile:

    (except it would be clearer if you used brackets: 2tanx/(1+tan^2x) …

    i suspect that's partly the reason why your next step was wrong …

    as Mark44 has already said, always use brackets (parentheses))

    NO!

    = 2tanx/sec2x …

    carry on from there :wink:
     
  16. Jun 11, 2013 #15
    Any idea on which identity to use from these point on? I'm lost :(
     
  17. Jun 11, 2013 #16

    CAF123

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    What is tanx equal to? What is sec2x equal to? Now sub in.
     
  18. Jun 11, 2013 #17
    =(2sinx/2cosx)/(1/cos^2x) ?
     
  19. Jun 11, 2013 #18

    tiny-tim

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    yes :smile:

    (except for one silly mistake :wink:)

    and then?​
     
  20. Jun 11, 2013 #19
    I have no idea, I feel likei cou,d multiply the top by 2cos leaving me 2sinx2cosx/cos which wou,d give me the answer? Correct?
     
  21. Jun 11, 2013 #20

    tiny-tim

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    tg22542, i don't understand your defeatism

    you say "i have no idea", even though you obviously do have an idea (and it's the correct one) :confused:

    yes, your 1/(1/something) can obviously be cancelled out

    it's your "2"s that need a second look :redface:
     
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