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Trigonometric Identites

  1. Dec 11, 2007 #1
    I need some help on 15 identities problems to help me study for my math final.

    They are blurry and somewhat hard to read, but if anyone wants to take a crack at some of them they are here:

    http://img407.imageshack.us/img407/5696/math1lh2.jpg

    http://img509.imageshack.us/img509/9642/math2fw6.jpg

    http://img70.imageshack.us/img70/7996/math3hn7.jpg

    http://img69.imageshack.us/img69/7543/dsc00180on3.jpg


    Thanks for any help on any of them you can give.
     
  2. jcsd
  3. Dec 11, 2007 #2
  4. Dec 11, 2007 #3
    I know that, and I wanted to say it there, but I wasn't sure if anyone would believe me, but this isn't homework, merely just something to help me study for my final on Thursday.
     
  5. Dec 11, 2007 #4

    Avodyne

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    Science Advisor

    We *still* want to see some attempts on your part to work them out.

    And, we don't want to read blurry gray-on-gray pages. Take the trouble to type them in, if you want us to type something in response.
     
  6. Dec 11, 2007 #5
    Ok, I'll ask more help oriented questions.

    The problem is:

    sin2t-cott = -cott cos2t

    if solving from the left side, does

    sin2t-cott = 2sintcost- (cost/sin)
     
  7. Dec 12, 2007 #6

    Avodyne

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    Yes. Now let's work on the right side. We have
    cos(2t)=cos^2(t)-sin^2(t)=2cos^2(t)-1=1-2sin^2(t)
    Which of these is likely to be the most useful? If you don't know, try all three; multiply each by -cot(t) = -cos(t)/sin(t) and see what you get.
     
  8. Dec 25, 2007 #7
    http://img69.imageshack.us/img69/7543/dsc00180on3.jpg

    For this one, you need to first figure out the sin 3x in terms of sin x (and cos3x in terms of cos x). Once you have done that, substitute them in and it will be rather easy.

    http://img407.imageshack.us/img407/5696/math1lh2.jpg

    Second one from the top, let 2theta = x and solve for x. very simple.

    On this one, third from the top. Remember what wonders you can do with cos^2 x and sin^2 x, especially on the left side. What two things multiply to give you cos^4 x?

    I think the rest are easy, just mention any other ones you are having difficulties with I don't wanna go through all, it's late :D
     
  9. Dec 25, 2007 #8

    rock.freak667

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    Homework Helper

    For this one I suggest you use the factor formulas as these are most helpful when given problems like this.

    these are the formulas to know:

    [tex]sinP + sinQ=2sin(\frac{P+Q}{2})cos(\frac{P-Q}{2})[/tex]

    and

    [tex]cosP + cosQ=2cos(\frac{P+Q}{2})cos(\frac{P-Q}{2})[/tex]
     
  10. Dec 25, 2007 #9
    actually it doesn't help in that question's case. finding sin3x is faster.
     
  11. Dec 25, 2007 #10

    rock.freak667

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    Homework Helper

    [tex]\frac{sin3x+sinx}{cos3x+cosx}

    =\frac{2sin2xcosx}{2cos2xcosx}=tan2x[/tex]
     
    Last edited: Dec 25, 2007
  12. Dec 27, 2007 #11
    I highly doubt if you are 'preparing' for your exam'. Sorry for the bluntness, but most of these are just 2 to 3 step solutions.

    I think you need to know some of the identities. Some are mentioned above, here are a few more

    sin (2t)= 2sin(t) cos (t)
    1-cos(t)= 2 sin^2(t/2)

    You just need to be familiar with the identities and then when you see it later on in life, you'll automatically know what to do.
     
    Last edited: Dec 27, 2007
  13. Dec 27, 2007 #12

    Dr Transport

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    Gold Member

    The easiest way to attack these problems is to convert ALL the quantities into sines and cosines before doing anything, simplify them from there and rearrange. Many of these will be solved in less than 10 lines.
     
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