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Trig Problems

  1. Aug 10, 2007 #1
    I know that sin 320 deg is equal to -sin 40 deg. However, how do you know that -sin 40 deg is equal to - cos 50 deg? Thanks
     
  2. jcsd
  3. Aug 10, 2007 #2
    Hmm... I think it is something about cofunctions. I'm going to read more into it first. =]
     
  4. Aug 10, 2007 #3
    You know because of the cofunction identities:

    [tex]\sin{\theta} = \cos{(90-\theta)}[/tex]

    [tex]\cos{\theta} = \sin{(90-\theta)}[/tex]

    [tex]\tan{\theta} = \cot{(90-\theta)}[/tex]

    [tex]\cot{\theta} = \tan{(90-\theta)}[/tex]

    [tex]\sec{\theta} = \csc{(90-\theta)}[/tex]

    [tex]\csc{\theta} = \sec{(90-\theta)}[/tex]
     
  5. Aug 10, 2007 #4
    Draw a right triangle and look at the relationships.
     
  6. Aug 10, 2007 #5
    Ok thanks.

    Also, how do you solve this problem?:

    [tex]\frac{x^{2}-9x+14}{3x^{3}-6x^{4}} \times \frac{2x-1}{x^{2}-2x-35}[/tex]

    I simplified it down to:

    [tex] \frac{(x-7)(x-2)}{3x^{3}(-2x+1)} \times \frac{2x-1}{(x-7)(x+5)}[/tex]

    Is there a way to cancel out (-2x+1) and 2x-1? Is it okay to multiply (-2x+1) by -1 or is it wrong because it changes the value? When I set them equal to one another they cancel out to equal 0. Why is this?

    Thanks.
     
  7. Aug 10, 2007 #6

    learningphysics

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    Homework Helper

    Don't set them equal... you can do a number of things... you can multiply the numerator and denominator by -1... that keeps the fraction the same, and lets you multiply out the (-2x+1) by -1...

    Or multiply the denominator by (-1)*(-1) which is just one so you're keeping the value of the fraction the same... but you can use one of the -1 to multiply (-2x+1) by -1...

    The essential idea is that -2x+1 = -1*(2x-1) = -(2x-1). You can also think of it as factoring out the -1.

    Remember that whatever operations you do, you don't want to change the value of the quantity you're evaluating...

    When you cancel things out with multiplication, you're dividing something by itself... so you don't get 0, but 1...
     
    Last edited: Aug 10, 2007
  8. Aug 10, 2007 #7
    Thanks for your reply! I didn't even see that you could factor -1 out. Does anyone know the correct definition of Standard Deviation and how to find it? I've searched on google and all that came up were some confusing equations...
     
  9. Aug 10, 2007 #8
    Yes, u have to use those confusing equations to calculate standard deviation.
    http://www.answers.com/topic/standard-deviation?cat=biz-fin#

    I liked the one that's in "Accounting Dictionary" section(scroll down to there), and
    if you further scroll down, wikipedia explains it all (including that bell curve)
     
  10. Aug 10, 2007 #9
    Okay, I'm nowhere near that level of math so I've found a defintion that I think I can apply. Please tell me if you think this is accurate or not:

    "The standard deviation is a measure of spread- how far the observations in a set of data are from their mean."

    Here's a problem of standard deviation:
    Of the following lists of numbers, which has the smallest standard deviation?
    a) 1, 5, 9
    b) 3, 5, 8
    c) 4, 5 ,8
    d)7, 8 , 9
    3) 8, 8, 8

    So would you calculate all of their means for example (7, 8, 9): (7+8+9)/3=8 and then see how far the data {7,8,9} are from that mean?

    In this case 8,8,8 has a mean of 8 which is the smallest out of all of them. Does that seem like the correct way to do it? Thanks
     
  11. Aug 11, 2007 #10

    Gib Z

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    Homework Helper

    Theres a certain method to calculate the standard deviation.

    First find the mean of each of those sets. Now Find the difference between each of the scores and the mean. Square each of them. Sum that, and divide that by the number of scores. Now square root the whole thing. That gives the Standard Deviation.
     
    Last edited: Aug 11, 2007
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