Results 1 to 2 of 2

Thread: Trig Triangle Proof

  1. #1
    Junior Member
    Joined
    Oct 2009
    Posts
    37

    Trig Triangle Proof

    Hello Everyone,

    I have another question again...

    In any acute-angle triangle ABC, prove that:
    b^2sin2C + c^2sin2B = 2bcsinA

    Does it have to do with sine law and cosine law? I can't figure it out.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by KelvinScale View Post
    Hello Everyone,

    I have another question again...

    In any acute-angle triangle ABC, prove that:
    b^2sin2C + c^2sin2B = 2bcsinA

    Does it have to do with sine law and cosine law? I can't figure it out.
    $\displaystyle b^2\sin{2C} + c^2\sin{2B} = 2b^2\sin{C}\cos{C} + 2c^2\sin{B}\cos{B}$


    Now from the cosine rule:

    $\displaystyle c^2 = a^2 + b^2 - 2ab\cos{C}$

    $\displaystyle 2ab\cos{C} = a^2 + b^2 - c^2$

    $\displaystyle \cos{C} = \frac{a^2 + b^2 - c^2}{2ab}$.


    Similarly $\displaystyle \cos{B} = \frac{a^2 + c^2 - b^2}{2ac}$.



    So $\displaystyle 2b^2\sin{C}\cos{C} + 2c^2\sin{B}\cos{B} = 2b^2\sin{C}\left(\frac{a^2 + b^2 - c^2}{2ab}\right) + 2c^2\sin{B}\left(\frac{a^2 + c^2 - b^2}{2ac}\right)$

    $\displaystyle = \frac{b(a^2 + b^2 - c^2)\sin{C}}{a} + \frac{c(a^2 + c^2 - b^2)\sin{B}}{a}$

    $\displaystyle = \frac{b(a^2 + b^2 - c^2)\sin{C} + c(a^2 + c^2 - b^2)\sin{B}}{a}$



    We also know from the sine rule:

    $\displaystyle \frac{\sin{C}}{c} = \frac{\sin{A}}{a}$

    so $\displaystyle \sin{C} = \frac{c\sin{A}}{a}$.

    Similarly $\displaystyle \sin{B} = \frac{b\sin{A}}{a}$.


    So $\displaystyle \frac{b(a^2 + b^2 - c^2)\sin{C} + c(a^2 + c^2 - b^2)\sin{B}}{a}$

    $\displaystyle = \frac{b(a^2 + b^2 - c^2)\left(\frac{c\sin{A}}{a}\right) + c(a^2 + c^2 - b^2)\left(\frac{b\sin{A}}{a}\right)}{a}$

    $\displaystyle = \frac{bc\sin{A}(a^2 + b^2 - c^2)}{a^2} + \frac{bc\sin{A}(a^2 + c^2 - b^2)}{a^2}$

    $\displaystyle = \frac{a^2bc\sin{A}}{a^2} + \frac{b^3c\sin{A}}{a^2} - \frac{bc^3\sin{A}}{a^2} + \frac{a^2bc\sin{A}}{a^2} + \frac{bc^3\sin{A}}{a^2} - \frac{b^3c\sin{A}}{a^2}$

    $\displaystyle = \frac{2a^2bc\sin{A}}{a^2}$

    $\displaystyle = 2bc\sin{A}$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Right triangle in trig
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: May 28th 2010, 04:24 PM
  2. Replies: 2
    Last Post: Dec 9th 2009, 05:09 AM
  3. Replies: 3
    Last Post: Apr 30th 2009, 07:41 AM
  4. triangle trig
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Oct 7th 2008, 06:20 PM
  5. triangle trig
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Jul 28th 2007, 09:35 AM

Search tags for this page

Search Tags


/mathhelpforum @mathhelpforum