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
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
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,
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
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.
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
Use this IDENTITY: [tex] \sin x\sin y\equiv \frac{1}{2}[\cos(x-y)-\cos(x+y)] [/tex] The result is immediate. Daniel.
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.
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.