Results 1 to 3 of 3

Thread: trig identity

  1. #1
    Junior Member
    Joined
    Dec 2008
    Posts
    57

    Exclamation trig identity

    Prove that:

    cos(2x-3y)+ cos3y = cot a
    sin(2x-3y)+ sin3y
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    849
    Hello, math123456!

    You have a typo . . .


    Prove that: .$\displaystyle \frac{\cos(2x-3y)+ \cos3y}{\sin(2x-3y) + \sin3y} \:=\:\cot{\color{red}x}$
    We need two Sum-to-Product identities:

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

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



    Numerator:
    $\displaystyle \cos(2x-3y)+\cos3y \:=\:2\cos\left(\frac{2x-3y+3y}{2}\right)\cos\left(\frac{2x-3y-3y}{2}\right) \:=\:2\cos x\cos(x-3y)$


    Denominator:
    $\displaystyle \sin(2x-3y)+\sin3y \:=\:2\sin\left(\frac{2x-3y+3y}{2}\right)\cos\left(\frac{2x-3y-3y}{2}\right) \:=\:2\sin x\cos(x - 3y)$


    The fraction becomes: .$\displaystyle \frac{2\,\cos x\,\cos(x - 3y)}{2\,\sin x\,\cos(x - 3y)} \;=\;\frac{\cos x}{\sin x} \;=\;\cot x$

    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by math123456 View Post
    Prove that:

    cos(2x-3y)+ cos3y = cot a
    sin(2x-3y)+ sin3y
    Use the identities

    $\displaystyle \sin{a} + \sin{b} = 2\sin{\frac{a+b}{2}}\cos{\frac{a-b}{2}}$

    and

    $\displaystyle \cos{a} + \cos{b} = 2\cos{\frac{a+b}{2}}\cos{\frac{a-b}{2}}$.


    So $\displaystyle \frac{\cos{2x - 3y} + \cos{3y}}{\sin{2x - 3y} + \sin{3y}}$

    $\displaystyle = \frac{2\cos{\frac{2x - 3y + 3y}{2}}\cos{\frac{2x - 3y - 3y}{2}}}{2\sin{\frac{2x - 3y + 3y}{2}}\cos{\frac{2x - 3y - 3y}{2}}}$

    $\displaystyle = \frac{\cos{x}\cos{(x - 3y)}}{\sin{x}\cos{(x - 3y)}}$

    $\displaystyle = \frac{\cos{x}}{\sin{x}}$

    $\displaystyle = \cot{x}$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. trig identity help...
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: May 17th 2011, 09:43 AM
  2. Another trig identity
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Feb 14th 2011, 04:54 PM
  3. Trig Identity
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Apr 18th 2010, 05:43 PM
  4. Trig identity
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Mar 9th 2010, 11:39 PM
  5. Odd Trig Identity
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Aug 27th 2008, 08:06 PM

Search Tags


/mathhelpforum @mathhelpforum