Results 1 to 10 of 10

Thread: Circle Theorem and Trig Equation Problem!

  1. #1
    Newbie
    Joined
    May 2008
    Posts
    16

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    May 2008
    Posts
    16

    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 =]
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    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 $\displaystyle \frac{\cos^2 x}{\sin x + \sin^2 x} = \frac{1 - \sin x}{\sin x}$. Then you will understand why the solutions to the equation $\displaystyle \frac{\cos^2 x}{\sin x + \sin^2 x} = 2$ are got by solving the following two equations:


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


    2. $\displaystyle 1 + \sin x = 0 \Rightarrow \sin x = -1$.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1
    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
    $\displaystyle \frac{\cos ^2 \theta}{\sin \theta + \sin ^2 \theta} = 2$

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

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

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

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

    $\displaystyle 1 - \sin \theta = 2 \sin \theta$

    $\displaystyle 1 = 3 \sin \theta $

    $\displaystyle \sin \theta = \frac 13$

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

    $\displaystyle \theta = 19.5^\circ, 161^\circ \ \ \text{to 3 sf}$
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    May 2008
    Posts
    16
    thanks air, you made the first problem clear
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    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 =]
    $\displaystyle x^2 + y^2 = 6x + 8y \Rightarrow (x^2 - 6x) + (y^2 - 8y) = 0 \Rightarrow (x - 3)^2 - 9 + (y - 4)^2 - 16 = 0$ .......
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    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: $\displaystyle \frac{\cos^2 x}{\sin x + \sin^2 x} = \frac{1 - \sin^2 x}{\sin x (1 + \sin x} = \frac{(1 - \sin x)(1 + \sin x)}{\sin x (1 + \sin x)}$ ......
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    May 2008
    Posts
    16
    ahhh i get it all now.

    thanks both of you

    i just missed the embarassingly simple maths =[











    thanks for the circle help too

    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1
    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 =]
    $\displaystyle x^2 + y^2 = 6x + 8y$

    $\displaystyle x^2 - 6x + y^2 - 8y = 0$

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

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

    $\displaystyle \therefore$ Radius of circle: $\displaystyle \sqrt{3^2 + 4^2 - 0} = 5 $.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    849
    Hello, acid_rainbow!

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

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

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

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

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


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

    And: .$\displaystyle 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} $

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Trig equation problem
    Posted in the Trigonometry Forum
    Replies: 8
    Last Post: Oct 17th 2011, 01:03 PM
  2. Trig word problem - solving a trig equation.
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: Mar 14th 2011, 07:07 AM
  3. trig equation problem
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Apr 21st 2009, 11:46 AM
  4. Trig Equation Problem
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: Mar 5th 2009, 11:37 AM
  5. Replies: 2
    Last Post: Nov 12th 2008, 07:38 PM

Search Tags


/mathhelpforum @mathhelpforum