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Trig identities

by nelraheb
Tags: identities, trig
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nelraheb
#1
Mar11-05, 07:25 PM
P: 6
In triangle ABC if sin (A/2) sin (B/2) sin (C/2) = 1/8
prove that the triangle is equilateral plz show steps
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James R
#2
Mar11-05, 11:11 PM
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Perhaps you can use the sine rule (?)
nelraheb
#3
Mar12-05, 04:59 AM
P: 6
Sure .. I tried but had no success .. If you find an answer plz post your steps

A_I_
#4
Mar12-05, 05:51 AM
P: 137
Trig identities

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
VietDao29
#5
Mar12-05, 08:33 AM
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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,
nelraheb
#6
Mar12-05, 11:34 AM
P: 6
For AI thank you but this won't do
nelraheb
#7
Mar12-05, 11:39 AM
P: 6
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
dextercioby
#8
Mar12-05, 12:04 PM
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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.
nelraheb
#9
Mar12-05, 02:33 PM
P: 6
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
dextercioby
#10
Mar12-05, 03:35 PM
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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.
Tom Mattson
#11
Mar12-05, 03:40 PM
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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.
nelraheb
#12
Mar12-05, 05:20 PM
P: 6
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
dextercioby
#13
Mar12-05, 05:25 PM
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Quote Quote by nelraheb
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
That's interesting.The checking part.I've said IDENTITY. 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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