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An exponential problem and a trig problem

  1. Apr 5, 2006 #1
    Yesterday in a math competition, I came across two problems that I couldn't (and still can't) figure out how to solve under the competion conditions (in under three minutes, without using a calculator).

    The first one involved expential functions. When I try to do it I just get a huge mess of exponents and logs that takes me forever to simplify. It is as follows:

    9^x + 9^-x = 34, evaluate 3^x+3^-x

    The second problem involved trig functions:

    Find the exact value of cot 15 + cot 75

    Are there sum and difference formulas for cot? I tried turning the cot into cos/sin and then adding the fractions. That left me with
    [tex] \frac{\sin(15)\cos(75)+\cos(15)\sin(75)}{\sin(75)\sin(15)} [/tex]
    Using the sum formula this would become
    [tex]\frac{\sin 90}{\sin(75)\sin(15)} = \frac{1}{\sin(75)\sin(15)} [/tex]
    Can I do anything with this?

    Thanks for your help!
    Tom
     
  2. jcsd
  3. Apr 5, 2006 #2
    for the first problem, notice that 9^x=3^(2x) since 9=3^2, so substitute that in, then factor.

    for the second problem 75= 60+15, and 15=30-15. now you can use the sum formulas for the denominator. i didn't do it out, but i'd try that.
     
  4. Apr 5, 2006 #3

    Integral

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    Another approach for the second problem is to see that

    15 = 45 - 30 and that 75 = 45 + 30
     
  5. Apr 5, 2006 #4
    Ok, I got the second one, it's 4. I was avoiding using the sum formulas because I thought that I would just get a large mess of sines and cosines, but if you use integral's approach, the sums up being the difference of two squares, and it isn't really that complicated.

    How can I factor 3^(2x)+3^(-2x)?
     
  6. Apr 5, 2006 #5
    yeah, integral's was a better way for using the sums, was a bit simpler, same idea. it does work the way i explained too.

    for the factoring, isn't there a 3^2 in both those terms? after that, its very simple.
     
  7. Apr 6, 2006 #6
    I don't understand how I can factor a 3^2 out.
    3^2(3^x+3^-x) is equal to (3^x)(3^2)+(3^-x)(3^2) or 3^(x+2)+3^(-x+2).
    Also, 3^(2x)/3^2 would just become 3^(2x-2).
     
  8. Apr 6, 2006 #7
    sorry. you're right, you can take out the two though. here:

    (3^x + 3^-x)^2 = 3^2x + 3^-2x +2(3^x)(3^-x)= 3^2x + 3^-2x +2(3^0)

    so
    3^2x + 3^-2x = (3^x + 3^-x)^2 - 2 = 34

    now it works out nicely eh?
     
  9. Apr 7, 2006 #8
    Wow, great. That does work out very nicely. That is a pretty tricky problem, but thats what I would expect from this competition.
    Thanks!
    Tom
     
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