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Thread: Trig indentity

  1. #1
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    Trig indentity

    Hi All,

    How about do I go to answer this trig. identity question. I have attached it as a picture as wasn't to sure on how to insert the symbols.

    Also if it is not possible to plot the sketch just explain it in words what I have to do, i'm sure I would be able to understand.

    Thanks.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by dadon
    Hi All,

    How about do I go to answer this trig. identity question. I have attached it as a picture as wasn't to sure on how to insert the symbols.

    Also if it is not possible to plot the sketch just explain it in words what I have to do, i'm sure I would be able to understand.

    Thanks.
    I can't show the graph, but I can help with the rest.

    Via the identity:
    $\displaystyle R cos(\theta - \beta)=R cos\theta cos\beta + R sin\theta sin\beta$

    By comparison then:
    $\displaystyle a cos\theta + b sin\theta = R cos\theta cos\beta + R sin\theta sin\beta$
    So
    $\displaystyle a = R cos\beta$
    $\displaystyle b = R sin\beta$

    Dividing these two equations gives us:
    $\displaystyle \frac{R sin\beta}{R cos\beta}=tan\beta = \frac{b}{a}$
    or
    $\displaystyle \beta = tan^{-1} \left ( \frac{b}{a} \right ) $

    and
    $\displaystyle R^2 sin^2 \beta + R^2 cos^2 \beta = R^2 (1) = a^2+b^2$
    so
    $\displaystyle R = \sqrt{a^2+b^2}$

    Now, in order to graph $\displaystyle 2 cos\theta + 3 sin\theta$ you are essentially graphing:
    $\displaystyle \sqrt{2^2+3^2}cos \left ( \theta - \left [ tan^{-1} \left ( \frac{3}{2} \right ) \right ] \right ) $

    or roughly $\displaystyle \sqrt{13} \, cos(\theta - 0.983)$ vs. $\displaystyle \theta$.
    This graph is simply a cosine graph, with amplitude $\displaystyle \sqrt{13}$ and translated 0.983 radians (or so) to the right.

    -Dan
    Last edited by topsquark; May 2nd 2006 at 09:16 AM.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by dadon
    Hi All,

    How about do I go to answer this trig. identity question. I have attached it as a picture as wasn't to sure on how to insert the symbols.

    Also if it is not possible to plot the sketch just explain it in words what I have to do, i'm sure I would be able to understand.

    Thanks.
    Here is the plot

    RonL
    Attached Thumbnails Attached Thumbnails Trig indentity-wire.jpg  
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  4. #4
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    Post re:

    Hey Cheers for that guys!

    I get want you have done throughout but how do you get $\displaystyle
    a = R cos\beta
    $ and $\displaystyle
    b = R sin\beta
    $
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by topsquark
    $\displaystyle a cos\theta + b sin\theta = R cos\theta cos\beta + R sin\theta sin\beta$
    You get them by comparing forms. Let me write it out a bit more explicitly:
    $\displaystyle a cos\theta + b sin\theta = R cos\theta cos\beta + R sin\theta sin\beta$

    Now, take a look at the $\displaystyle cos\theta$ term. They have to be the same on both sides of the equation so we can write:
    $\displaystyle a cos\theta = R cos\theta cos\beta$
    Dividing both sides by $\displaystyle cos\theta$ we obtain:
    $\displaystyle a = R cos\beta$

    A similar argument with the sine function gives the second relation.

    Technically, I am assuming that $\displaystyle cos\theta \neq 0$ so some care needs to be taken as far as interpretation for all values of $\displaystyle \theta$. However, since $\displaystyle \theta$ is being taken as an independent variable and not a constant I feel we can more or less disregard this problem. (Theoretically I suppose we should be plotting an open circle at any point that $\displaystyle cos\theta = 0$.)

    -Dan
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  6. #6
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    Post re:

    Thanks for sharing you time to solve my problem and explainly in a way that I understand fully.

    Cheers

    dadon
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