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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

    VietDao29

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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]
    to
    [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

    dextercioby

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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.

    Daniel.
     
  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

    dextercioby

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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.

    Daniel.
     
  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:

    A+B+C=180
    C=180-A-B

    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

    dextercioby

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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).

    Daniel.
     
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