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Trig Identities that I can't get a grip on

  1. May 31, 2009 #1
    [tan(pi/4+x)-tan(pi/4-x)]/[tan(pi/4+x)+tan(pi/4-x)]=2sinxcosx

    I tried to prove this trig identity but I an really stuck. I think tan of pi/4 is '1', and if I do that then my numerator becomes zero, thus zero=2sinxcosx. But that can't be right, so I don't know what to do now.


    LS= (1+tanx-1-tanx)/(1+tanx+1-tanx)

    I get 0=2sinxcosx

    Any thoughts? Also, I am wondering where this advanced formatting option is or how to do it.
     
  2. jcsd
  3. May 31, 2009 #2
    You will need to use the identities

    tan(A+B)=(tanA+tanB)/(1-tanA*tanB)

    tan(A-B)=(tanA-tanB)/(1+tanA*tanB)
     
  4. May 31, 2009 #3
    Thankyou Overt, I did not see it.

    So would I be right in assigning 'y' as pi/4? and let 'x' be 'x'?

    For tan(pi/4+x) I will instead get (tanx+tanpi/4)/(1-tanxtanpi/4)?
     
  5. May 31, 2009 #4
    Correct!
     
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