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Math Help - Circle Theorem and Trig Equation Problem!

  1. #1
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    Circle Theorem and Trig Equation Problem!

    Hey i have two problems that i am stuck on

    1) Solve the following quation for 0 < theta < 180

    (i will use x for theta, i know how to solve using the ASTC graph however i cannot rearrange this)

    Cos^2(x) / Sinx + Sin^2(x)= 2

    We have been told earlier that this is equal to 1-sinx/sinx
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  2. #2
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    Problem # 2

    Circle Problem

    Show that the centre of the circle with equation:

    x^2 + y^2 = 6x + 8y is (3,4) and find the radius


    Thanks =]
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  3. #3
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    Quote Originally Posted by acid_rainbow View Post
    Hey i have two problems that i am stuck on

    1) Solve the following quation for 0 < theta < 180

    (i will use x for theta, i know how to solve using the ASTC graph however i cannot rearrange this)

    Cos^2(x) / Sinx + Sin^2(x)= 2

    We have been told earlier that this is equal to 1-sinx/sinx
    You should make sure you know why \frac{\cos^2 x}{\sin x + \sin^2 x} = \frac{1 - \sin x}{\sin x}. Then you will understand why the solutions to the equation \frac{\cos^2 x}{\sin x + \sin^2 x} = 2 are got by solving the following two equations:


    1. \frac{1 - \sin x}{\sin x} = 2 \Rightarrow 1 - \sin x = 2 \sin x \Rightarrow \sin x = \frac{1}{3}.


    2. 1 + \sin x = 0 \Rightarrow \sin x = -1.
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  4. #4
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    Quote Originally Posted by acid_rainbow View Post
    Hey i have two problems that i am stuck on

    1) Solve the following quation for 0 < theta < 180

    (i will use x for theta, i know how to solve using the ASTC graph however i cannot rearrange this)

    Cos^2(x) / Sinx + Sin^2(x)= 2

    We have been told earlier that this is equal to 1-sinx/sinx
    \frac{\cos ^2 \theta}{\sin \theta + \sin ^2 \theta} = 2

    \frac{1 - \sin ^2 \theta}{\sin \theta + \sin ^2 \theta} = 2

    \frac{1 - \sin ^2 \theta}{\sin \theta ( 1 + \sin  \theta)} = 2

    \frac{(1 - \sin \theta)(1 + \sin \theta)}{\sin \theta ( 1 + \sin  \theta)} = 2

    \frac{1 - \sin \theta}{\sin \theta } = 2

    1 - \sin \theta = 2 \sin \theta

    1  = 3 \sin \theta

    \sin \theta = \frac 13

    \theta = \sin ^{-1} \left( \frac 13 \right)

    \theta = 19.5^\circ, 161^\circ \ \ \text{to 3 sf}
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  5. #5
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    thanks air, you made the first problem clear
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  6. #6
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    Quote Originally Posted by acid_rainbow View Post
    Circle Problem

    Show that the centre of the circle with equation:

    x^2 + y^2 = 6x + 8y is (3,4) and find the radius


    Thanks =]
    x^2 + y^2 = 6x + 8y \Rightarrow (x^2 - 6x) + (y^2 - 8y) = 0 \Rightarrow (x - 3)^2 - 9 + (y - 4)^2 - 16 = 0 .......
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  7. #7
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    Quote Originally Posted by acid_rainbow View Post
    i still dont understand.

    i tried simplifying it and ended up geting -sinx = 2 which is impossible
    What don't you understand? What is the it that you tried simplifying and ended up getting -sinx = 2? Have you read my earlier post?

    Note: \frac{\cos^2 x}{\sin x + \sin^2 x} = \frac{1 - \sin^2 x}{\sin x (1 + \sin x} = \frac{(1 - \sin x)(1 + \sin x)}{\sin x (1 + \sin x)} ......
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  8. #8
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    ahhh i get it all now.

    thanks both of you

    i just missed the embarassingly simple maths =[











    thanks for the circle help too

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  9. #9
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    Quote Originally Posted by acid_rainbow View Post
    Circle Problem

    Show that the centre of the circle with equation:

    x^2 + y^2 = 6x + 8y is (3,4) and find the radius


    Thanks =]
     x^2 + y^2 = 6x + 8y

     x^2 - 6x + y^2 - 8y = 0

    A general form is x^2 + y^2 + 2gx + 2fy + c = 0 where center of circle: (-g, -f) and the radius of circle: \sqrt{g^2 + f^2 - c}.

    \therefore Center of circle: \left(-\left(\frac{-6}{2}\right), -\left(\frac{-8}{2}\right)\right) = (3, 4).

    \therefore Radius of circle: \sqrt{3^2 + 4^2 - 0} = 5 .
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  10. #10
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    Hello, acid_rainbow!

    1) Solve the following equation for 0^o \leq \theta \leq 180^o

    Cos^2(x) / Sinx + Sin^2(x) = 2 .
    . . . What is in the denominator?

    We have been told earlier that this is equal to \frac{1-\sin x}{\sin x} .
    . . . Ah, this helps!

    We have: . \frac{\cos^2\!x}{\sin x + \sin^2\!x} \:=\:2 \quad\Rightarrow\quad 1 - \sin^2\!x \;=\;2(\sin x + \sin^2\!x)

    which simplifies to: . 3\sin^2\!x + 2\sin x - 1 \:=\:0\quad\Rightarrow\quad (\sin x + 1)(3\sin x - 1) \:=\:0


    Then: . \sin x + 1 \:=\:0\quad\Rightarrow\quad\sin x \:=\:-1\quad\Rightarrow\quad x \:=\:270^o \quad\hdots not in the interval

    And: . 3\sin x -1\:=\:0\quad\Rightarrow\quad\sin x \:=\:\frac{1}{3}\quad\Rightarrow\quad\boxed{ x \;\approx\;19.47^o,\;160.53^o}

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