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Trig identities

  1. Mar 11, 2005 #1
    Trig identities plz help

    In triangle ABC if sin (A/2) sin (B/2) sin (C/2) = 1/8
    prove that the triangle is equilateral plz show steps
  2. jcsd
  3. Mar 11, 2005 #2

    James R

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    Perhaps you can use the sine rule (?)
  4. Mar 12, 2005 #3
    Sure .. I tried but had no success .. If you find an answer plz post your steps
  5. Mar 12, 2005 #4
    one method suggested:

    (an absurd reasonning)

    if it is an equilateral triangle then:
    A = B = C = pi/3 rad

    implies ---> A/2 = B/2 = C/2 = pi/6 rad

    implies ---> sin(A/2) = sin(B/2) = sin(C/2) = 1/2

    implies ---> sin(A/2)sin(B/2)sin(C/2) = 1/2*1/2*1/2 = 1/8

    thus it is indeed an equilateral triangle

    if i come with another one i will post it :)
    hope it will help
  6. Mar 12, 2005 #5


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    Try expand the equation
    [tex]\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}} = \frac{1}{8}[/tex]
    [tex]4\sin{\frac{C}{2}}^{2} - 4\sin{\frac{C}{2}}\cos{\frac{A - B}{2}} + 1 = 0[/tex]
    Then to
    [tex](2\sin{\frac{C}{2}} - \cos{\frac{A - B}{2}})^{2} + (\sin{\frac{A - B}{2}})^{2} = 0[/tex]
    Now you have something like [itex]A^{2} + B^{2} = 0[/itex] so
    [tex]\left\{ \begin{array}{c} A = B \\ \sin{\frac{C}{2}} = \frac{1}{2}\cos{\frac{A - B}{2}} \end{array}\right[/tex]
    So you will have A = B = C = 60 degrees, which implies the triangle ABC is equilateral.
    Hope it help.
    Viet Dao,
  7. Mar 12, 2005 #6
    For AI thank you but this won't do
  8. Mar 12, 2005 #7
    For VietDao29 If A^2 + B^2 = 0 then we're stuck because no two +ve numbers addto zero ...right ? Then it should be A^2 = - B^2
    How did you expand 1st step
    How did you get last step
    plz go in more details
  9. Mar 12, 2005 #8


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    Both A and B are real.So their square is larger or equal to zero.In order for the sum of the squares to be 0,each if the squares must be 0.

  10. Mar 12, 2005 #9
    Well that's a good point. How did I miss that :) Now for the first step plz how did we expand Sin (A/2) Sin (B/2) Sin (C/2) to next step
    ie. How to start .... the rest is ok
  11. Mar 12, 2005 #10


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    Use this IDENTITY:

    [tex] \sin x\sin y\equiv \frac{1}{2}[\cos(x-y)-\cos(x+y)] [/tex]

    The result is immediate.

  12. Mar 12, 2005 #11

    Tom Mattson

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    Try starting by eliminating a variable. Since you know that A, B, and C are all in the same triangle, you have:


    See where that gets you.
  13. Mar 12, 2005 #12
    Thank you all ... I can do it now following your steps
    The rule supplied by Dextercioby did not look familiar (but it's correct I checked) ..well memory is not what it used to be :) isn't that a bit complicated though ... I thought the answer should be more straight forward .. any way thank you all again
  14. Mar 12, 2005 #13


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    That's interesting.The checking part.I've said IDENTITY. :wink: There may have been a chance i didn't invent it,but either picked it from a book or deduced starting other identities (which i have actually done).

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