Results 1 to 8 of 8

Math Help - Challenging trigonometry problem

  1. #1
    Newbie
    Joined
    Feb 2011
    Posts
    7

    Challenging trigonometry problem

    Please, I need help to prove:

    (\sin A + \sin B \sin C)^2 \leq  1 + \sin A + \sin B \sin C
    such that A, B, and C are triangle angles

    Thank you
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Have you expanded out the LHS? Can you apply any identities here?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2011
    Posts
    7
    I tried many expansions! I just could not prove it...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by mestmoha View Post
    I tried many expansions! I just could not prove it...
    Please post all your working.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Feb 2011
    Posts
    7

    Challenging trigonometry problem

    we should prove:
    \alpha = 1 + \sin A + \sin B \sin C - (\sin A + \sin B \sin C)^2 \geq 0
    we know that:

    \frac{\sin A}{a} = \frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{2R}
    such that R is the cimscribed circle radius

    by Substituting in alpha, we need to prove now:

    \frac{16 R^4 + 8 aR^3 + 4 bc R^2 - (4 a^2 R^2 + 4 abc R + b^2 c^2) } {16 R^4} \geq 0

    Such that a, b,and c are the edges of a triangle.

    As a remark,

    R=\frac{abc}{\sqrt{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}}

    Thank you,
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Feb 2011
    Posts
    68
    Quote Originally Posted by mestmoha View Post
    we should prove:
    \alpha = 1 + \sin A + \sin B \sin C - (\sin A + \sin B \sin C)^2 \geq 0
    Where did alpha come from? Did you forget to write it out in your first post?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Feb 2011
    Posts
    7
    Alpha does not change anything, it is just the difference between the LHS and the RHS
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Feb 2011
    Posts
    7
    Quote Originally Posted by mestmoha View Post
    Alpha does not change anything, it is just the difference between the LHS and the RHS
    I think I found the solution. Please, check it for me:
    \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}
    \frac{{\sin}^2 A}{{a}^2}=\frac{\sin B \sin C}{bc}
    \sin A + \sin B \sin C= \Sin A (1 + \frac{bc}{{a}^2} \sin A)

    For any A, and a \sin A + \sin B \sin C is maximum if and only if b=c (you can use the area= 1/2 bc sin A to approach it).

    hence for a triangle with b=c:

    \sin A + \sin B \sin C= \sin A + \frac{1}{2} \cos A +\frac{1}{2}

    if:
    x=\sin A   \quad then \quad f(x) =x + \frac{1}{2}\sqrt{1-x^2} + \frac {1} {2}
    by using the derrivative of f(x), and setting f'(x)=0 we get:
    x=\sin A= \frac{2}{\sqrt{5}}

    Then the maximum value of f(x)   \sin A + \sin B \sin C \quad ) is \frac{1+\sqrt{5}}{2}

    now we proved that for a triangle ABC \sin A + \sin B \ sin C \leq \frac{1 + \sqrt{5}}{2}

    Let's consider a function g(x) such that:
    g(x)=-x^2 + x +1

    The roots of g(x) are : \frac{1-\sqrt{5}}{2}  \quad and \quad \frac{1+\sqrt{5}}<br />
{2}

    Hence:

    for \quad every \quad  x \quad: \frac{1-\sqrt{5}}{2} \leq x \leq \frac{1+\sqrt{5}}<br />
{2}

    and since :

    0\leq \sin A + \sin B \sin C \leq \frac {1+\sqrt{5}}{2}  (as we proved)

    Then:

    g(\sin A + \sin B \sin C) \geq 0

    -->

    - {(\sin A + \sin B \sin C)}^{2} + (\sin A + \sin B \sin C) +1 \geq 0

    Thus :

    (\sin A + \sin B \sin C)}^{2} \leq  (\sin A + \sin B \sin C) +1
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Challenging Problem
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: November 22nd 2009, 11:18 AM
  2. Challenging Problem
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: August 15th 2009, 09:53 AM
  3. Challenging problem
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: July 13th 2009, 05:49 AM
  4. challenging sequences problem
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: February 26th 2009, 04:41 PM
  5. Challenging Trigonometry Question
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: June 11th 2007, 04:08 PM

Search Tags


/mathhelpforum @mathhelpforum