Results 1 to 8 of 8

Math Help - questions about trigo and circles

  1. #1
    Junior Member
    Joined
    Jun 2008
    Posts
    53

    questions about trigo and circles



    don't even know how to begin, does it involve factor/sum formulae?



    angle ABC and angle ADC is 90 deg respectively. because AC is diameter.
    angle BAD is 180deg-θ. because of cyclic quadrilateral.....
    how do i relate them and find AD?




    sin x cos x = -(3/10)
    then ....can i do this :

    sin x = - (3/10) , cos x = - (3/10)

    ??? or i can only do this when it's equal to zero?





    what i tried to do:

    centre(0,0)
    thus, i roughly know it would be something like (x^2)+(y^2).....
    A(a,0)
    B(0,b)
    AB=2sqrt2

    so..

    sqrt{[(a-0)^2]+[(0-b)^2]}=2sqrt2
    (a^2)+(b^2)=8

    and then......i'm stuck ="=
    how am i going to find the radius ...
    and the question never mention about radius/diameter...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2008
    Posts
    792
    1. Use the identities:
    \sin{A}-\sin{B} = 2\cos{\left(\frac{A+B}{2}\right)}\sin{\left(\frac{  A-B}{2}\right)}
    \cos{A}+\cos{B}=2\cos{\left(\frac{A+B}{2}\right)}\  cos{\left(\frac{A-B}{2}\right)}

    to verify that [1]/[2] is indeed the identity in question.

    ===========

    [1] \sin{x}-\sin{y} = 2\cos{\left(\frac{x+y}{2}\right)}\sin{\left(\frac{  x-y}{2}\right)}
    [2] \cos{x} + \cos{y} = 2\cos{\left(\frac{x+y}{2}\right)}\cos{\left(\frac{  x-y}{2}\right)}<br />
    \implies \frac{2\cos{\left(\frac{x+y}{2}\right)}\sin{\left(  \frac{x-y}{2}\right)}}{2\cos{\left(\frac{x+y}{2}\right)}\c  os{\left(\frac{x-y}{2}\right)}} = \frac{\sin{\left(\frac{x-y}{2}\right)}}{\cos{\left(\frac{x-y}{2}\right)}} = \tan{\left(\frac{x-y}{2}\right)}

    Surely you know what to do next?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,245
    Thanks
    1
    1) \sin x-\sin y=2\sin\frac{x-y}{2}\cos\frac{x+y}{2}
    \cos x+\cos y=2\cos\frac{x-y}{2}\cos\frac{x+y}{2}

    2) Let \widehat{BCA}=\alpha, \ \widehat{ACD}=\beta
    AC=\sqrt{AB^2+BC^2}=\sqrt{85}
    \sin\alpha=\frac{AB}{AC}=\frac{7}{\sqrt{85}}, \ \cos\alpha=\frac{BC}{AC}=\frac{6}{\sqrt{85}}

    Now, \sin\beta=\sin(\theta-\alpha)=\sin\theta\cos\alpha-\sin\alpha\cos\theta=\frac{6}{\sqrt{85}}\sin\theta-\frac{7}{\sqrt{85}}\cos\theta
    But \sin\beta=\frac{AD}{AC}=\frac{AD}{\sqrt{85}}

    Now, find AD.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,245
    Thanks
    1
    3) 10\sin x\cos x=5\cdot 2\sin x\cos x=5\sin 2x
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,245
    Thanks
    1
    A(a,0), \ B(0,a)
    Try again and find a.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,548
    Thanks
    541
    Hello, wintersoltice!

    Given: . \begin{array}{cccc}{\color{blue}[1]} &\sin x - \sin y &=&\frac{1}{2} \\ \\[-4mm] {\color{blue}[2]} & \cos x + \cos y &=& \frac{1}{4} \end{array} . find the value of: . \tan\left(\tfrac{x-y}{2}\right)

    . . \text{(i) }\;\tfrac{1}{8}\qquad\text{(ii) }\;0 \qquad\text{(iii) }\;\tfrac{1}{2}\qquad\text{(iv) }\;2
    We need two Sum-to-Product identities:

    . . \sin A - \sin B \:=\:2\cos\left(\tfrac{A+B}{2}\right)\sin\left(\tf  rac{A-B}{2}\right)

    . . \cos A + \cos B \:=\:2\cos\left(\tfrac{A+B}{2}\right)\cos\left(\tf  rac{A-B}{2}\right)


    Then [1] and [2] become: . \begin{array}{ccccc}<br />
2\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right) &=& \frac{1}{2} & {\color{blue}[3]} \\ \\[-3mm]<br />
2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right) &=& \frac{1}{4} & {\color{blue}[4]} \end{array}


    Divide [3] by [4]: . \frac{2\cos\left(\frac{x+y}{2}\right)\sin\left(\fr  ac{x-y}{2}\right)}{2\cos\left(\frac{x+y}{2}\right)\cos\  left(\frac{x-y}{2}\right)} \;=\;\frac{\frac{1}{2}}{\frac{1}{4}} . \Rightarrow\quad \frac{\sin\left(\frac{x-y}{2}\right)}{\cos\left(\frac{x-y}{2}\right)} \;=\;2

    . . Therefore: . \tan\left(\frac{x-y}{2}\right) \:=\:2 . . . answer (iv)


    Edit: Too slow again . . .
    .
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Jun 2008
    Posts
    53
    Quote Originally Posted by red_dog View Post
    3) 10\sin x\cos x=5\cdot 2\sin x\cos x=5\sin 2x



    10 sin x cos x +3 =0
    (5)(2)sin x cos x =-3
    5 sin 2x =-3
    sin 2x = -(3/5)
    basic angle of 2x = 36.87 deg
    90deg < x < 180 deg .......=> second quadrant
    therefore,
    2x= (180deg - 36.87deg)
    2x= 143.13 deg
    x= 71.565 deg

    >"<....
    there's no this answer among the choices.....
    where did i went wrong ?? @"@
    based on my sixth sense, i feel that the part i higlighted in bold is incorrect/weird/etc(don't know how to describe)..........


    Quote Originally Posted by red_dog View Post
    A(a,0), \ B(0,a)
    Try again and find a.
    why you know they are the same???
    i mean the unknown, a.
    Last edited by wintersoltice; December 31st 2008 at 05:02 AM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,245
    Thanks
    1
    Quote Originally Posted by wintersoltice View Post
    1

    why you know they are the same???
    i mean the unknown, a.
    Because OA and OB are radius in the circle, so OA=OB.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: July 29th 2010, 11:32 AM
  2. Need help with 2 trigo questions
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: June 18th 2009, 07:06 AM
  3. questions about differentiation of trigo
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 11th 2009, 05:23 AM
  4. Help! 2 questions - triangles and circles
    Posted in the Geometry Forum
    Replies: 1
    Last Post: November 1st 2008, 10:48 AM
  5. Questions on trigo. equation
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: September 17th 2007, 07:01 PM

Search Tags


/mathhelpforum @mathhelpforum