Question

PT:sin a/2 + sin b/2 + sin c/2 -1 = 4sin((π-a)/4) .sin( (π-b)/4) .sin ((π-c)/4 )

If a+b+c=180

Answer 1

sin a/2  + sin b/2 + sin c/2  -1      
{apply  formula on first two terms sinC + sinD = 2sin((C+D)/2)cos((C-D)/2)}
 
=>   2 sin((a+b)/4) cos((a-b)/4) + sin c/2   - 1
{u ca write 1 as sin π/2 }

=> 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)

Hence Proved

Answer 2

Answer:

The solution is in attachment above.