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Homework Help: Prove this trigonometric identity

  1. Aug 12, 2008 #1
    1. The problem statement, all variables and given/known data
    Verify the possibility of an identity graphically. (Completed this part)
    Then, prove each identity algebraically.


    2. Relevant equations





    3. The attempt at a solution








    That's as far as I got and now I have no idea what to do.
  2. jcsd
  3. Aug 12, 2008 #2
    You went off track on the first step.

    [tex] \frac{sinx+ \frac{sinx}{cosx}}{cosx+1} \neq \frac{sinx+sinxcosx}{cosx+1}[/tex]

    This identity is looks straightforward. Just add the terms in the numerator correctly.
  4. Aug 12, 2008 #3
    What do you mean by add the numerator? I see how I went wrong but how does the math work in that?
  5. Aug 12, 2008 #4


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    From here:


    Multiply both the numerator and denominator by cosx. Your numerator of sinx+sinxcosx is correct. But the denominator is incorrect,
  6. Aug 12, 2008 #5
    A super easy trick is to notice this

    sin(x)(1+ 1/cos(x)) / cos(x)(1+ 1/cos(x)) = sin(x)/cos(x) =? hmmm.....factoring helps
  7. Aug 12, 2008 #6


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    Multiply both sides by (cos(x)+1). Now you have sin(x)+tan(x)=tan(x)*(1+cos(x)). Multiply out the right side.
    Last edited: Aug 12, 2008
  8. Aug 12, 2008 #7
    Your fundamental approach is sort of off. Instead of just working with the left side to try to make it a tangent (which is hard), you should be operating on both sides to try to get sides that are clearly equal.
  9. Aug 12, 2008 #8
    I'm just trying to learn by how the textbook shows, and that's the only way they've shown so far.

    I tried out the way Dick showed (it worked) and I'm going to keep trying the other methods as well, thanks.
  10. Aug 12, 2008 #9
    Easiest way is to put everything in terms of sin & cos:

    sin(x)+tan(x) / cos(x)+1 = tan(x) =>

    sin(x)+(sin(x)/cos(x)) / cos(x)+1 = sin(x)/cos(x) =>replace tan and multiply by denominator; then expand

    sin(x)+(sin(x)/cos(x)) = sin(x)+(sin(x)/cos(x)) =>can factor and simplify from here

    Hope that helps,
  11. Aug 12, 2008 #10


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    They are all really the same method. Just manipulate the thing algebraically until you you get an identity. As Alex6200 said, don't restrict yourself to freezing one side. It's often easier if you mix them up.
  12. Aug 15, 2008 #11
    Hmm this is the one thing about verifying trigonometric identities that's confused me a bit. In precalc we were always told to start with one side and try to get the expression on the other side. But most of the identities were already true, so it really doesn't matter if you simplify expressions on both sides does it? I mean any other operation besides restricting oneself to a simplification on one side is just the basic principle of applying the same algebraic manipulation to both sides.

    I mean the whole point is to get you to use the fundamental identities in a creative way. And often if not always you could leave one side alone (usually the less complicated looking side) and manipulate the other side enough to get two equal expressions. I think at least for some identities this is more challenging.
  13. Aug 16, 2008 #12
    I have a new question, I'm trying to work through this sheet of similar questions but it's just frustrating. I have tried simplifying this one question down so many ways and it's not equal and there isn't an answer in the back of the book for it so I'm wondering if it's not equal? Here it is:

    Prove the result algebraically:


    I would show my work but I've tried so many times it will just be frustrating.
  14. Aug 16, 2008 #13


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    Start by factoring. a^2-b^2=(a-b)(a+b). Apply with a=sin^2 and b=cos^2. Do you see anything to do from there?
  15. Aug 16, 2008 #14
    Work with the LHS

    Step 1. Factor it, you should get a product of 2 things

    Step 2. What does each product = ? (use the fact that [tex] sin^{2}(x) \, + \, cos^{2}(x) \, = \, 1 [/tex] for both parts and you will get the RHS.
  16. Aug 16, 2008 #15


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    Notice that the left-hand side is a difference of two squares. If you factor it accordingly, you will notice something about one of the factors that will let you simplify the left side considerably. The remaining factor can then be easily manipulated (using a familiar trig identity) to look like the right-hand side.
  17. Aug 16, 2008 #16
    Oh ok that makes so much sense now, thank you.
  18. Aug 16, 2008 #17
    My general approach there would be to see that you have sines and cosines, remember that sine and cosine are related by the Pythagorean theorem, and then get it all into sines or cosines (in this case sine is easier).

    Then you have a straightforward problem.
  19. Aug 16, 2008 #18
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