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

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

Grade:11

1 Answers

kkbisht
90 Points
5 years ago
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

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