Proving an Equilateral Triangle: Trig Identities Explained

[nelraheb]

Trig identities please help

In triangle ABC if sin (A/2) sin (B/2) sin (C/2) = 1/8
prove that the triangle is equilateral please show steps

 

[James R]

Perhaps you can use the sine rule (?)

 

[nelraheb]

Sure .. I tried but had no success .. If you find an answer please post your steps

 

[A_I_]

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]

Try expand the equation
\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}} = \frac{1}{8}
to
4\sin{\frac{C}{2}}^{2} - 4\sin{\frac{C}{2}}\cos{\frac{A - B}{2}} + 1 = 0
Then to
(2\sin{\frac{C}{2}} - \cos{\frac{A - B}{2}})^{2} + (\sin{\frac{A - B}{2}})^{2} = 0
Now you have something like A^{2} + B^{2} = 0 so
\left\{ \begin{array}{c} A = B \\ \sin{\frac{C}{2}} = \frac{1}{2}\cos{\frac{A - B}{2}} \end{array}\right
So you will have A = B = C = 60 degrees, which implies the triangle ABC is equilateral.
Hope it help.
Viet Dao,

 

[nelraheb]

For AI thank you but this won't do

 

[nelraheb]

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
please go in more details

 

[dextercioby]

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]

Well that's a good point. How did I miss that :) Now for the first step please 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]

Use this IDENTITY:

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

The result is immediate.

Daniel.

 

[quantumdude]

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]

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]

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.