Results 1 to 5 of 5

Math Help - solving cosine ratio equations

  1. #1
    jrr
    jrr is offline
    Newbie
    Joined
    Feb 2010
    Posts
    5

    solving cosine ratio equations

    I have an identily of the form:

    cos(a1 b)/cos(a2 b) =(x1/x2)*cos(a1)/cos(a2)

    Given a1,a2,x1 and x2, how can I solve it for b?

    Thanks in advance for any help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello jrr

    Welcome to Math Help Forum!
    Quote Originally Posted by jrr View Post
    I have an identily of the form:

    cos(a1 b)/cos(a2 b) =(x1/x2)*cos(a1)/cos(a2)

    Given a1,a2,x1 and x2, how can I solve it for b?

    Thanks in advance for any help
    It's not clear what the symbols are between the \cos and the brackets, so I shall ignore them for now. (If it turns out to be a power, like \cos^2(a_1-b) the same method applies.)

    Expand each of the cosine expressions on the LHS using \cos(A-B) = \cos A \cos B + \sin A\sin B, and then divide top and bottom by \cos b. Then solve the resulting equation for \tan b.

    Here's the start:
    \frac{\cos(a_1-b)}{\cos(a_2-b)}=\frac{x_1}{x_2}\cdot\frac{\cos a_1}{\cos a_2}

    \Rightarrow\frac{\cos a_1\cos b+\sin a_1\sin b}{\cos a_2\cos b+\sin a_2\sin b}=\frac{x_1}{x_2}\cdot\frac{\cos a_1}{\cos a_2}

    \Rightarrow\frac{\cos a_1+\sin a_1\tan b}{\cos a_2+\sin a_2\tan b}=\frac{x_1}{x_2}\cdot\frac{\cos a_1}{\cos a_2}
    Can you now make \tan b the subject of this equation, and hence find b?

    Grandad
    Follow Math Help Forum on Facebook and Google+

  3. #3
    jrr
    jrr is offline
    Newbie
    Joined
    Feb 2010
    Posts
    5

    Thanks

    Thanks Grandad,
    That makes it starightforward now

    jrr
    Follow Math Help Forum on Facebook and Google+

  4. #4
    jrr
    jrr is offline
    Newbie
    Joined
    Feb 2010
    Posts
    5
    I re-arranged to get

    tan(b)= (x1-x2).cos(a1).cos(a2)/(x2.cos(a2).sin(a1)-x1.cos(a1).sin(a2))

    but I was wondering if this could be simplified any further?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello jrr
    Quote Originally Posted by jrr View Post
    I re-arranged to get

    tan(b)= (x1-x2).cos(a1).cos(a2)/(x2.cos(a2).sin(a1)-x1.cos(a1).sin(a2))

    but I was wondering if this could be simplified any further?
    Yes, you can divide top-and-bottom by \cos a_1\cos a_2 to get:
    \tan b = \frac{x_1-x_2}{x_2\tan a_1 - x_1\tan a_2}
    Grandad
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. integral of ratio of cosine and polynomial
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 21st 2011, 01:48 PM
  2. Solving a cosine integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 6th 2011, 04:41 PM
  3. Ratio Test for Taylor Expansion of Cosine
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 7th 2010, 09:43 PM
  4. Replies: 8
    Last Post: January 6th 2010, 07:23 PM
  5. Sine/Cosine Equations
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: February 21st 2009, 02:17 AM

Search Tags


/mathhelpforum @mathhelpforum