sina/2+sinb/2+sinc/2=1+4sinpie-a/4.sinpie-b/4.sin

Question

sina/2+sinb/2+sinc/2=1+4sinpie-a/4.sinpie-b/4.sinpie-c/4

kkbisht · Accepted Answer

You have not mentioned the condition : a+b+c= $\pi$ =180

LHS= Sin a/2 + sin b/2 + sin c/2 -1
(Now usr T-identity sinC + sinD = 2sin(C+D)/2)cos(C-D)/2)

=> 2 sin((a+b)/4) cos((a-b)/4) + sin c/2 -1

=> 2 sin ((a+b)/4) cos((a-b)/4) + sin c/2 -sin π/2

{again apply formula on last two terms sinC - sinD = 2 cosA sinB }

=> 2sin ((a+b)/4) cos((a-b)/4) + 2cos((π+c)/4)sin((c-π)/4)
{now a+b+c = π so a+b=π-c}

=> 2sin((π-c)/4) cos((a-b)/4) + 2cos((π+c)/4) sin(( c-π)/4)

=>2sin((π-c)/4){ cos((a-b)/4) - cos((π+c)/4)}
{now again use formula cos(A-B) - cos(A+B) = 2 sin((C+D)/2) sin ((D-C)/2)}

=> 2sin((π-c)/4){ 2 sin ((a-b+π+c)/8) sin((π+c-a+b)/8)}

{now use a+b+c=π in the last two terms}

=>2sin((π-c)/4) { 2 sin ((2π-2b)/8) sin (( 2π- 2a)/8)}
now after simplify all calculation, we will get

=> 4 sin ((π-c)/4) sin((π-b)/4)sin((π-a)/4)

therefore Sin a/2 + sin b/2 + sin c/2 -1=4 sin ((π-c)/4) sin((π-b)/4)sin((π-a)/4)

Hence Proved

By

kkbisht

sina/2+sinb/2+sinc/2=1+4sinpie-a/4.sinpie-b/4.sinpie-c/4