sin^2 a/2+ sin^2b/2+ sin^2c/2= 1-2sina/2sinb/2sinc/2 prove?

Answer 1

Write left hand side statement one side and leave 1 1/2 page then write Right hand side statement and then apply formula from L.H.S start writing down words and in same manner from R.H.S. and start writing up words.
I want you to tell the secret you can write ant thing in between when don't find formula's could be applied anymore.

Happy??
L.H.S. = R.H.S.

proved.

Answer 2

here is the solution:
=sin^2 a/2+ sin^2b/2+ sin^2c/2
=1 - [ Cos^2 a/2 - Sin^2 b/2 ] + Sin^2 c/2
=1 - Cos (a+b)/2 X Cos(a-b)/2 + Sin^2 c/2 [ a+b+c= π]
= 1- Sin c/2 cos (a-b)/2 + Sin^2 c/2
=1- Sin c/2 [ cos (a-b)/2 - Sin c/2 ]
= 1 - Sin c/2 [ cos (a-b)/2 - Cos (π/2 - c/2 ) ]
=1 - Sin c/2 [ 2 Sin a/2 Sin b/2 ]
=1 - 2 Sin a/2 Sin b/2 Sin c/2
Hence Proved if any doubts ask me.
There is no mistakes in this, but i have removed some steps from this answer as they were very simple.

Answer 3

hey! im sorry to tell u that this term is wrong!
suppose a/2 = b/2 = c/2 = x
then:

sin^2(x) + sin^2(x) + sin^2(x) = 1 - 2sin(x)sin(x)sin(x)
=> 3sin^2(x) = 1 - 2sin^3(x)
=> 3sin^2(x) + 2sin^3(x) = 1

now just let x = pi/3

=> 3 ( (3^(1/2))/2 )^2 + 2 ((3^(1/2))/2)^3 = 1
=> 3 ( 3/4 ) + 2 ( 3( 3^(1/2) )/8 ) = 1
=> 9/4 + (3*(3 ^ (1/2))/4 = 1
=> 9 + 3(3 ^ (1/2)) = 4
and yes,it is inconsistent!

Answer 4

this is based on the properties of triangle, which should have been clearly mentioned in the question. In such a case a+b+c= 180 deg or a/2 +b/2+c/2 =90deg

Answer 5

this is confusing ....do yo mean (1-2sina)/(2sinb)/(2sinc)/2or
1-[(2sina)/2]*[sinb/2]*[sinc/2]
or [(1-2sina)/2]*[sinb/2]*[sinc/2]

Answer 6

hay man/woman . can u explain what is SIN.didi not learn this concept till now

Answer 7

is the problem correctly given?