Trig identities

1. nelraheb

6
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. James R

562
Perhaps you can use the sine rule (?)

3. nelraheb

6
Sure .. I tried but had no success .. If you find an answer plz post your steps

4. A_I_

137
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

5. VietDao29

1,422
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,

6. nelraheb

6
For AI thank you but this won't do

7. nelraheb

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

8. dextercioby

12,304
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.

9. nelraheb

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

10. dextercioby

12,304
Use this IDENTITY:

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

The result is immediate.

Daniel.

11. Tom Mattson

5,539
Staff Emeritus
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.

12. nelraheb

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

13. dextercioby

12,304
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.