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Simplifying Trigonometric Expressions

  1. Aug 31, 2009 #1
    1. The problem statement, all variables and given/known data
    I'm not sure if you can ask about two questions in one thread; they are very similar. If it's not allowed I can make a second thread and edit this one.

    Simplify the expressions:

    1. sin^3 theta + cos^3 theta / sin theta + cos theta

    2. 2-tan theta / 2 csc theta-sec theta


    2. Relevant equations
    Pythagorean Identities


    3. The attempt at a solution

    1. if the either of these were squared, they could be equal to 1. I thought that I could get rid of th denominator and make sin^2 and cos^2 that way...and it would simplify to 1. The answer I am given is actually 1-sin theta cos theta. I think this has to factor somehow, but I'm not certain. I'm not used to dealing with problems that look like this yet.

    2. I think this has to do with factoring something out as well, but I'm not sure what. I tried changing them all to variations of sin and cos, but nothing clicked for me.
     
  2. jcsd
  3. Aug 31, 2009 #2
    I wonder how they've got that answer, I don't see anything wrong with what you did. :confused:

    What happens if we use the identities

    [tex]sin^2 \theta = \frac{1}{2} (1-cos2 \theta)[/tex]

    [tex]cos^2 \theta = \frac{1}{2} (1+cos2 \theta)[/tex]
     
  4. Aug 31, 2009 #3

    rock.freak667

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    Homework Helper

    For

    [tex]\frac{sin^3 \theta + cos^3 \theta}{sin \theta + cos \theta}[/tex]

    you should notice that sin3θ+cos3θ=0 when sinθ=cosθ, so try factoring the numerator as (sinθ+cosθ)*f(θ)
     
  5. Aug 31, 2009 #4
    I'm sorry, I don't understand. Why would sin3theta+cos3theta=0 when sin theta=cos theta. Maybe it's something we haven't covered yet?

    We are expected to get the answer from the even odd properties and the Pythagorean Identities.

    I apologize if I'm making this difficult. :(
     
  6. Aug 31, 2009 #5
    My recommendation for the first problem is to ignore momentarily sine and cosine and focus on how one would simplify

    [tex]\frac{a^3+b^3}{a+b}[/tex]

    and then deal with the resulting expression.

    As for the second problem I am assuming you are working with the expression

    [tex]\frac{2-\tan \theta}{2 \csc \theta - \sec \theta}.[/tex]

    If not, please correct me. If so, then I'd recommend multiplying the numerator and denominator by [itex]\sin \theta \cdot \cos \theta[/itex] and see if that helps simplify the ratio.

    --Elucidus
     
  7. Aug 31, 2009 #6

    rock.freak667

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    Homework Helper

    I meant that you can factor a3+b3 into (a+b)*(something).
     
  8. Aug 31, 2009 #7
    A quick reply for you Elucidus on the first problem. THANK YOU! I get it now. It's the formula (x+y)(x^2-xy+y^2) only with sin and cos. I worked it out on some paper and it fell into place. It's so frustrating when I can't see ways to solve the problem that I've already learned.

    As for the second problem, you have the expression correct. I'm going to work on it now.
     
  9. Aug 31, 2009 #8
    OK, I've worked through the second problem. I doubted myself on this one. I changed everything to sin/cos and thought about multiplying everything by sin and cos. I thought that wasn't right and I wasn't sure if I was multiplying correctly anyway. I should just try my ideas next time.

    Anyway, it all fell into place and I understand how it works. Now I can sleep well tonight. Thank you all so much for your time and patience. :)
     
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