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Prove the trigonometric identity

  1. Mar 11, 2013 #1
    1. The problem statement, all variables and given/known data
    (question attached)


    2. Relevant equations



    3. The attempt at a solution
    Checking solution.. pretty sure I did this wrong.
    (solution attached)
     

    Attached Files:

  2. jcsd
  3. Mar 11, 2013 #2

    Mark44

    Staff: Mentor

    Yes, you did. I see several mistakes.

    On the left side, tanθ - 1 = sinθ/cosθ - 1, but that's not equal to (sinθ - 1)/cosθ, as you show.

    On the right side there are two mistakes that I see: you factored the denominator wrong in the next to the last line, and I have no idea how you ended with what you did in the last line.
     
  4. Mar 11, 2013 #3

    Mark44

    Staff: Mentor

    I would recommend working with the right side, to make it look like the left side. This can be done in three steps.
     
  5. Mar 12, 2013 #4
    Oh wow okay, I think I got it. Thank you very much
     

    Attached Files:

  6. Mar 12, 2013 #5
    Sorry nevermind, I see what you were saying earlier now. So the last step on the right side would be to simplify it further to tanθ -1
     
  7. Mar 12, 2013 #6

    Mark44

    Staff: Mentor


    No, you're not getting it at all. On the right side, you have this:
    $$ \frac{sin^2(θ) - cos^2(θ)}{sin(θ)cos(θ) + cos^2(θ)} = \frac{sin^2(θ) - 1}{sin(θ)cos(θ)}$$

    This is wrong. It appears that you are cancelling the cos2(θ) terms to arrive at the expression on the right, above. This is like saying (5 + 1)/(2 - 1) = (5 - 1)/2 = 2. The actual value is 6.

    The only time you can cancel is when the same factor (not term) appears in the numerator and denominator. You will not be successful in trig until you get a stronger grasp on algebra.
     
    Last edited: Mar 12, 2013
  8. Mar 13, 2013 #7
    No matter what I do I can't get the answer. I don't understand.
     
  9. Mar 13, 2013 #8

    Mark44

    Staff: Mentor

    Do you know how to factor a2 - b2?
    Can you factor ab + b2?

    You need to know how to make these factorizations so that you can simplify this:
    $$ \frac{sin^2(θ) - cos^2(θ)}{sin(θ)cos(θ) + cos^2(θ)} $$

    The way you did it is not valid.
     
  10. Mar 13, 2013 #9
    So the numerator would be (sinƟ+cosƟ)(sinƟ+cosƟ)
    But I don't know how to factor ab + b^2
     
  11. Mar 13, 2013 #10
    Oh! ab + b^2 would equal b(a+b)?
    So the denominator would be cosƟ(sinƟ + cosƟ)?
     
  12. Mar 13, 2013 #11

    Mark44

    Staff: Mentor

    It's because you do not understand how to simplify fractions and rational expressions.

    In post #1 you did this:

    $$ \frac{sin^2(θ) - (1 - sin^2(θ))}{sin(θ) cos(θ) (1 - sin^2(θ))} ~~\text{(1)}$$
    $$ \frac{sin^2(θ) - 1}{sin(θ) cos(θ)} ~~\text{(2)}$$
    $$ \frac{sin(θ) - 1 }{cos(θ)} ~~\text{(3)}$$

    In (1) you have the wrong sign in the denominator.
    In (2), you apparently cancelled (1 - sin2(θ)). You absolutely can't do this, since what you cancelled is not a factor of both numerator and denominator.
    In (3), you apparently cancelled sin(θ), which is also invalid since sin(θ) is not a factor of the numerator.

    In post #4 you did this:
    $$ \frac{sin^2(θ) - cos^2(θ)}{sin(θ) cos(θ) + cos^2(θ)} ~~\text{(4)}$$
    $$ \frac{sin^2(θ) - 1)}{sin(θ) cos(θ) } ~~\text{(5)}$$
    $$ \frac{sin(θ) - 1}{cos(θ) } ~~\text{(6)}$$

    In (4), you have the sign in the denominator correct this time.
    In (5), you cancelled cos2(θ), which is not valid for the reasons stated above.
    In (6), you sin(θ), which is also not valid. sin(θ) is not a factor in the numerator.
     
  13. Mar 13, 2013 #12

    Mark44

    Staff: Mentor

    If you multiply this back out do you get sin2(θ) - cos2(θ)?



    Yes!
     
  14. Mar 13, 2013 #13
    Okay so now I have (sinƟ+cosƟ)(sinƟ+cosƟ)/cosƟ(sinƟ + cosƟ)
    So now I can actually cancel out (sinƟ + cosƟ) once on the top and once on the bottom, to get
    sinƟ + cosƟ/cosƟ

    From here I'm stuck again. I know the sinƟ/cosƟ would equal tan, but I'm not sure where the -1 comes from
     
  15. Mar 13, 2013 #14

    Mark44

    Staff: Mentor

    Does (sinƟ+cosƟ)(sinƟ+cosƟ) multiply out to sin2Ɵ - cos2Ɵ?

     
  16. Mar 13, 2013 #15
    Ah sorry I had the right answer in my head but I wrote it out wrong. Didn't even notice.
    (sinƟ+cosƟ)(sinƟ-cosƟ)/cosƟ(sinƟ + cosƟ)

    Then sinƟ-cosƟ/cosƟ
    Now I'm stuck again. I'm doing different variations of substituting the quotient identity into the equation but I can't get the right order
     
  17. Mar 13, 2013 #16

    Mark44

    Staff: Mentor

    So far, so good, but you need some more parentheses around the entire denominator, like so:
    (sinƟ+cosƟ)(sinƟ-cosƟ)/(cosƟ(sinƟ + cosƟ))
    Here you need some parentheses around the entire numerator, like so:

    (sinƟ-cosƟ)/cosƟ
    (A + B)/C is the same as A/C + B/C, and (A - B)/C is the same as A/C - B/C.
    You're having a lot of difficulties working with fraction-type expressions. It would be a good idea for you to take some time reviewing how to simplify fractions and rational expressions. If you don't still have the textbook you studied from, there are a number of online sites that are helpful, such as http://khanacademy.org.
     
  18. Mar 13, 2013 #17
    Got it! sinƟ/cosƟ - cosƟ/cosƟ
    = tanƟ-1

    Ahh thank you so much. I definitely need a refresher it's been a long time since I've taken a math course. Thank you for the link and for you patience and help!!!!
     
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