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Proving identities

  1. Aug 14, 2016 #1
    1. The problem statement, all variables and given/known data
    prove the following identity

    upload_2016-8-14_2-22-59.png

    2. Relevant equations
    no equations required
    3. The attempt at a solution
    I've been trying to prove this identity, but no matter what I do, I can't seem to make both sides the same
    here is my answer to this qts so far: can someone please tell me what I have to do next?
    Im pretty sure i'm supposed to be getting tan(x) - 1 on both sides. So, should I leave the left side the same (keep it as tan(x) -1) and work only with the right side, until both sides are equal?
    any help would be appreciated. Thanks.
    upload_2016-8-14_2-25-13.png


    Sorry for accidently posting the same thread twice :p
     

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  2. jcsd
  3. Aug 14, 2016 #2

    ehild

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    When you divide a sum or difference, you have to divide all terms. ##\frac{\sin(θ)-\cos(θ)}{\cos(θ)}## is not sin(θ)-1. And you made the same error in the right side.
    Leave the left side as it is, and work with the right side. Divide both numerator and denominator with cos2(θ), for example.
     
  4. Aug 14, 2016 #3
    Thanks for the quick reply!
    Ok so i understand that i am supposed to leave the left side the same, but what do i do with the right side. You mentioned dividing both numerator and denominator with cos^2(θ), how do i do that?
    so is it:
    sin^2(x) - cos^2(x) /cos^2(x)
    -------------------------------------
    sin(x) cos(x) + cos^2(x) / cos^2(x)

    i was originally using trigonometric identities:
    for ex: sin^2(x) + cos^2(x) = 1 so i was assuming that sin^2(x) - cos^2(x) = -1
    and so i replaced sin^2(x) - cos^2(x) with -1
    is this correct?
     
  5. Aug 14, 2016 #4

    ehild

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    No. You have to divide the whole expressions, not only the last terms. Do not omit the parentheses.
    No. It is wrong. sin^2(x) - cos^2(x) is not -1. Try with numbers. 2/3 + 1/3 = 1, is it true, that 2/3 - 1/3 = -1???
     
  6. Aug 14, 2016 #5

    Math_QED

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    sin^2 x + cos^2 x = 1
    => -sin^2 x - cos^2 x = -1 and not what you wrote.

    Leave the left side as it is. In the right side, we notice that we can use the formula: a^2 - b^2 = (a-b)(a+b) in the numerator. Try to find the factors of the denumerator and the problem becomes quite trivial then.
     
  7. Aug 14, 2016 #6
    ok so for the right side,if I use the formula: a^2 - b^2 = (a-b)(a+b)
    then it becomes:

    (sin(x) - cos(x)) (sin(x) + cos(x))
    -------------------------------------
    sin(x) cos(x) + cos^2(x)

    how do i solve this??
     
  8. Aug 14, 2016 #7

    dextercioby

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    You're not getting the right push so far. You need to show that a = b/c. Thus ac = b. Cross-multiply (tan theta - 1) by the denominator (sin theta cos theta + cos^2 theta) theta and simplify.
     
  9. Aug 14, 2016 #8
    srry im very confused, what do you mean by cross multiply and why would I need to cross multiply tan(x) - 1 by the denominator of the right side?
    Can you provide an example!
     
  10. Aug 14, 2016 #9
    ok so i came up with an answer, can someone tell me if im on the right track?

    ((sin^2(x) - cos^2(x)) /cos^2(x)
    -------------------------------------
    (sin(x) cos(x) + cos^2(x)) / cos^2(x)

    = tan^2(x)
    ----------
    tan(x) + 1

    = tan(x) tan(x)
    -------------
    tan(x) + 1

    = tan(x) + 1

    im not sure what to do from here.
     
  11. Aug 14, 2016 #10

    ehild

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    The last equation is wrong.
     
  12. Aug 14, 2016 #11
    can you please elaborate?
     
  13. Aug 15, 2016 #12

    SammyS

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    Look at that numerator again !

    What is ##\displaystyle \frac{\sin^2(x) - \cos^2(x)}{\cos^2(x)} \ ## ?

    It's not
    You may need to review: adding and subtracting fractions !
     
  14. Aug 15, 2016 #13
    for the numerator:

    sin^2(x) - cos^2(x) / cos^2 (x)
    inorder to find the answer, I have to divide cos^2(x) by each term in the expression:
    so sin^2(x) / cos^2(x) = tan^2(x) (since sin(x)/cos(x) = tan(x))
    and -cos^2(x) /cos^2(x) = -1

    so is the answer tan^2(x) - 1
    ???
     
  15. Aug 15, 2016 #14

    Math_QED

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    Factor out a cosx in the denumerator:

    [sin^2 (x) - cos^2(x)]/[sinxcosx + cos^2 x]
    = [(sinx - cosx)(sinx + cosx)]/[cosx(sinx + cosx)]

    Then use (a+b)/c = a/c + b/c
     
  16. Aug 15, 2016 #15

    SammyS

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    Yes.

    So make that correction in the following.
    That gives

    ## \displaystyle \frac{\tan^2(x) - 1}{\tan(x)+1} \ ##
     
  17. Aug 15, 2016 #16
    ((sin^2(x) - cos^2(x)) /cos^2(x)
    -------------------------------------
    (sin(x) cos(x) + cos^2(x)) / cos^2(x)


    = tan^2(x) - 1
    --------------
    tan(x) + 1

    = tan(x) - 1

    is this correct?
     
  18. Aug 15, 2016 #17

    SammyS

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    That is correct.

    Can you justify that last step ?
     
  19. Aug 15, 2016 #18
    ok so,

    tan^2(x) - 1
    ---------------
    tan(x) + 1

    = tan(x) tan(x) - 1
    ------------------ ( im pretty sure tan(x) crosses out at this part)
    tan(x) + 1

    = tan(x) - 1 / 1

    = tan(x) - 1

    is this process correct?
     
  20. Aug 16, 2016 #19

    SammyS

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    That's what I suspected.

    That is not the correct way to obtain the result.

    a2 -1 factors to (a - 1)(a + 1)..

    So, can you factor ##\ \tan^2(x) - 1 \ ?##
     
  21. Aug 16, 2016 #20
    ok so,

    (tan(x) - 1) (tan(x) + 1) (cross out the expression tan(x) + 1)
    -----------------------------
    tan(x) + 1

    = tan(x) - 1
     
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