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Thread: Differentiation and trigonometry question

  1. #1
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    Differentiation and trigonometry question

    A curve has parametric equations

    $\displaystyle x=\theta + sin\theta , y = 1 + cos\theta, (0 \leq \theta \leq \pi)$.

    Show that

    $\displaystyle (\frac{dx}{d\theta})^2 + (\frac{dy}{d\theta})^2 = 4cos^2(\frac{\theta}{2})$

    I differentiate the parametric equations, square them and get

    $\displaystyle cos^2\theta +2cos\theta + 1 + sin^2\theta$, and I know that somehow I need to show that it's equal to $\displaystyle 4cos^2(\frac{\theta}{2}$ but I have no idea - I've looked at some basic trigonometric identities to simplify it but I get nowhere. Is there anyone who can help me with this?

    Thanks if you have the time
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  2. #2
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    Quote Originally Posted by db5vry View Post
    A curve has parametric equations

    $\displaystyle x=\theta + sin\theta , y = 1 + cos\theta, (0 \leq \theta \leq \pi)$.

    Show that

    $\displaystyle (\frac{dx}{d\theta})^2 + (\frac{dy}{d\theta})^2 = 4cos^2(\frac{\theta}{2})$

    I differentiate the parametric equations, square them and get

    $\displaystyle cos^2\theta +2cos\theta + 1 + sin^2\theta$, and I know that somehow I need to show that it's equal to $\displaystyle 4cos^2(\frac{\theta}{2}$ but I have no idea - I've looked at some basic trigonometric identities to simplify it but I get nowhere. Is there anyone who can help me with this?

    Thanks if you have the time
    Since $\displaystyle sin^2(\theta)+cos^2(\theta)=1$, then :
    $\displaystyle cos^2(\theta) +2cos(\theta) + 1 + sin^2(\theta)=2cos(\theta)+2=2( \, cos(\theta) + 1 \, )$.

    Do you think you can do something if you know that $\displaystyle cos(\theta)=cos( \, 2 \, \frac{\theta}{2} \, )$ ?
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  3. #3
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    Quote Originally Posted by Miss View Post
    Since $\displaystyle sin^2(\theta)+cos^2(\theta)=1$, then :
    $\displaystyle cos^2(\theta) +2cos(\theta) + 1 + sin^2(\theta)=2cos(\theta)+2=2( \, cos(\theta) + 1 \, )$.

    Do you think you can do something if you know that $\displaystyle cos(\theta)=cos( \, 2 \, \frac{\theta}{2} \, )$ ?
    I didn't think of writing it in that form! Well following from your instructions I got

    $\displaystyle 2(cos(\frac{2\theta}{2}) + 1)$

    $\displaystyle cos^2\theta = \frac{1}{2} + cos2\theta$

    $\displaystyle 2cos^2\theta = 1 + 2cos2\theta$

    but from this I end up with $\displaystyle 4cos^2\theta$ which isn't the same as $\displaystyle 4cos^2(\frac{\theta}{2})$. Any ideas?
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  4. #4
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    Quote Originally Posted by db5vry View Post
    I didn't think of writing it in that form! Well following from your instructions I got

    $\displaystyle 2(cos(\frac{2\theta}{2}) + 1)$

    $\displaystyle cos^2\theta = \frac{1}{2} + cos2\theta$

    $\displaystyle 2cos^2\theta = 1 + 2cos2\theta$

    but from this I end up with $\displaystyle 4cos^2\theta$ which isn't the same as $\displaystyle 4cos^2(\frac{\theta}{2})$. Any ideas?
    Nope.
    Replace $\displaystyle cos( \, 2 \, \frac{\theta}{2} \, )$ by $\displaystyle 2cos^2( \, \frac{\theta}{2} \, ) - 1$.
    This comes from the identity $\displaystyle cos(2x)=2cos^2(x)-1$
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  5. #5
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    Quote Originally Posted by Miss View Post
    Nope.
    Replace $\displaystyle cos( \, 2 \, \frac{\theta}{2} \, )$ by $\displaystyle 2cos^2( \, \frac{\theta}{2} \, ) - 1$.
    This comes from the identity $\displaystyle cos(2x)=2cos^2(x)-1$
    I see! I was trying to derive it from going another way around which didn't work!

    So with $\displaystyle 2(cos\theta + 1)$

    and $\displaystyle cos2\theta = 2cos^2\theta - 1$

    it gives $\displaystyle 2(2cos^2(\frac{\theta}{2}))$ where the -1 and 1 cancel, therefore, we get

    $\displaystyle 4cos^2(\frac{\theta}{2}))$ which is what I was looking for!

    Thanks so much. That was really useful
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