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Thread: Trig equations need help(tried to work these but not sure if they are right)

  1. #1
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    Trig equations need help(tried to work these but not sure if they are right)

    Complete the form of the equation in rectangular form
    "r= -14 sin theta"

    I'm having factoring issues with this one, I got " x^2+(y+7)^2=14 but, when I try to check it, x^2+(y^2+14y+14)=14 doesn't come out the same. What am I doing wrong in the factoring?



    The rectangular coordinates of a point are given. Find the polar coordinates(r,theta) of this point with theta expresed in radians. Let r>0 and 0<or equal to theta <2pie
    (3,-3{squareroot}3)

    I used r=sqrt x^2 + y^2 to get 6
    then used tan inverse y/x to get -3sqrt3/3= -pie/3
    and got (6,-pie/3)
    did I do this correctly?
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  2. #2
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    Quote Originally Posted by ConfusedMath View Post
    Complete the form of the equation in rectangular form
    "r= -14 sin theta"

    I'm having factoring issues with this one, I got " x^2+(y+7)^2=14 but, when I try to check it, x^2+(y^2+14y+14)=14 doesn't come out the same. What am I doing wrong in the factoring?
    You have $\displaystyle r = -14\sin{\theta}$.


    You should know that $\displaystyle y = r\sin{\theta}$ by definition.

    So $\displaystyle \sin{\theta} = \frac{y}{r}$.


    Therefore:

    $\displaystyle r = -14\left(\frac{y}{r}\right)$

    $\displaystyle r = -\frac{14y}{r}$

    $\displaystyle r^2 = -14y$.


    You should also know that $\displaystyle x^2 + y^2 = r^2$.

    So $\displaystyle x^2 + y^2 = -14y$

    $\displaystyle x^2 + y^2 + 14y = 0$

    $\displaystyle x^2 + y^2 + 14y + 7^2 = 7^2$

    $\displaystyle (x - 0)^2 + (y + 7)^2 = 49$.
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  3. #3
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    Quote Originally Posted by ConfusedMath View Post
    The rectangular coordinates of a point are given. Find the polar coordinates(r,theta) of this point with theta expresed in radians. Let r>0 and 0<or equal to theta <2pie
    (3,-3{squareroot}3)

    I used r=sqrt x^2 + y^2 to get 6
    then used tan inverse y/x to get -3sqrt3/3= -pie/3
    and got (6,-pie/3)
    did I do this correctly?
    You have $\displaystyle (x, y) = (3, -3\sqrt{3})$.


    You know that $\displaystyle r^2 = x^2 + y^2$

    $\displaystyle r^2 = 3^2 + (-3\sqrt{3})^2$

    $\displaystyle r^2 = 9 + 27$

    $\displaystyle r^2 = 36$

    $\displaystyle r = 6$.


    Since $\displaystyle x = r\cos{\theta}$ and $\displaystyle y = r\sin{\theta}$

    This means

    $\displaystyle 6\cos{\theta} = 3$ and $\displaystyle 6\sin{\theta} = -3\sqrt{3}$.

    Since the cosine is positive and the sine is negative, this suggests the angle is in the fourth quadrant.

    Now, if we remember that

    $\displaystyle \frac{y}{x} = \frac{r\sin{\theta}}{r\cos{\theta}} = \tan{\theta}$


    $\displaystyle \frac{-3\sqrt{3}}{3} = \tan{\theta}$

    $\displaystyle \tan{\theta} = -\sqrt{3}$

    $\displaystyle \theta = 2\pi - \frac{\pi}{3}$, since $\displaystyle \theta$ is in the fourth quadrant.

    $\displaystyle \theta = \frac{5\pi}{3}$.
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    Thank you! That helps alot!
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