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Trig function products

  1. Dec 23, 2004 #1
    what exactly does [tex]\cos \theta_1 \cos \theta_2[/tex] represent, in relation to the angles? is this a dot product? i have played, and don't really see what this product is supposed to represent.

    EDIT: you know, i may have just answered my own question with a mere trig identity... perhaps i will figure this thing out i'm working on pretty soon.....
    Last edited: Dec 23, 2004
  2. jcsd
  3. Dec 24, 2004 #2

    matt grime

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    It would depend upon where these angles came from (dot product of what vectors?).

    Imagine I asked why x^2 represented - a slightly ambiguous question, isn't it?
  4. Dec 24, 2004 #3


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    A simple geometrical interpretation I can think of: if you angles t1 and t2 are contiguous, and at the same vertex, the factor cos(t1)cos(t2) is the factor that a segment would shrink upon succesive perpendicular projections through those angles.

    But, as Matt says, it would be useful to know more about the particular situation in which you are trying to interpret the product.
  5. Dec 24, 2004 #4
  6. Dec 24, 2004 #5


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    You refer to my post??What is ambiguous about that??It's true i didn't make any picture,but i relied upon your imagination.I guess u're all to familiar with spherical coordinates and the angles [itex] \theta [/itex] and [itex] \phi [/itex].Else,it's just math.A bit of ingenuity,though,else the geometric-triginomotric approach would have been more difficult to understand and would have necessitated a drawing.
    That is just a simple product of trigonometrical functions.It has no meaning other than the one specified already.If u don't like that product (though i still cannot imagine the reasons),u can use this formula to transform it into a sum of cosines:
    [tex] \cos \alpha \cos \beta=\frac{1}{2}[\cos(\alpha+\beta)+\cos(\alpha-\beta)] [/tex]

    I sincerely hope u're giving your best shot...Math is not a domain in which being lazy gives results...

  7. Dec 26, 2004 #6
    that's what i missed. i (embarrassingly) just didn't think of trig identities. btw, i posted this thread before you had replied to my problem. i do thank you for your well given response, all is understood now :-)
    Last edited: Dec 26, 2004
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