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Thread: trig differentiation

  1. #1
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    trig differentiation

    Hey i have this equation;
    a + b - 2abCos Ѳ = c + d - 2cdCos
    and i need to show that dѲ/d = cd sin / ab sinѲ

    im not quite sure how to do this
    i think that the 1st one goes to 2a + 2b + 2absinѲ
    and the 2nd one goes to 2c + 2d + 2cdsin

    but im stuck from there
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  2. #2
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    Quote Originally Posted by Jampop View Post
    Hey i have this equation;
    a + b - 2abCos Ѳ = c + d - 2cdCos
    and i need to show that dѲ/d = cd sin / ab sinѲ

    im not quite sure how to do this
    i think that the 1st one goes to 2a + 2b + 2absinѲ
    and the 2nd one goes to 2c + 2d + 2cdsin

    but im stuck from there
    This problem would have been better off in the calculus section. You're not in a precalculus class are you? If so, this is a difficult problem.

    Let $\displaystyle y=a^2+b^2-2abcos\theta$

    Therefore we also have:

    $\displaystyle y=c^2+d^2-2cdcos\phi$

    $\displaystyle \frac{d\theta}{d\phi}=\frac{d\theta}{dy}\frac{dy}{ d\phi}$

    Since $\displaystyle \frac{dy}{d\theta}=2absin\theta$, this means that $\displaystyle \frac{d\theta}{dy}=\frac{1}{2absin\theta}$.

    $\displaystyle \frac{dy}{d\phi}=2cdcos\phi$

    Substituting these derivatives into $\displaystyle \frac{d\theta}{d\phi}=\frac{d\theta}{dy}\frac{dy}{ d\phi}$ gives $\displaystyle \frac{d\theta}{d\phi}=\frac{cdsin\phi}{absin\theta }$
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  3. #3
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    Oh ok i understand how you get it now but can you just help me with a few things?

    Like how does



    Why it is not 2a + 2b + 2absinѲ??
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  4. #4
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    Quote Originally Posted by Jampop View Post
    Oh ok i understand how you get it now but can you just help me with a few things?

    Like how does



    Why it is not 2a + 2b + 2absinѲ??
    The numbers $\displaystyle a^2$ and $\displaystyle b^2$ are just constants. The derivative of a constant is zero. $\displaystyle \frac{d}{d\theta}a^2=0$. The same is true for $\displaystyle b^2$. Remember that a derivative is a rate of change with respect to some variable. In this case the variable is $\displaystyle \theta$. So the only part of the equation $\displaystyle a^2+b^2-2abcos\theta$ that changes with respect to $\displaystyle \theta$ is $\displaystyle -2abcos\theta$. The constants don't change (by definition of a constant), so their derivatives are zero.
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  5. #5
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    Oh yeah thanks very muchh
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