Identities.

I was unable to prove Q.2 and Q.5...

1) To Prove: if A+B+C=180°,
sin(B+2C) + sin(C+2A) + sin(A+2B) = 4 . sin(B-C)/2 . sin(C-A)/2 . sin(A-B)/2...

R.H.S.
= 4 . sin(B-C)/2 . sin(C-A)/2 . sin(A-B)/2
= [2sin(A-B)/2.sin(B-C)/2] x 2sin(C-A)/2
= [cos (A+C-2B)/2 - cos(A-C)/2] x 2sin(C-A)/2
= [2 . cos (A+C-2B)/2 . sin (C-A)/2] - [2 . cos(A-C)/2 . sin(C-A)/2]
= sin(C-B) - sin(A-B) + 2cos(A-C)/2 . sin(A-C)/2
= -sin(B-C) - sin(A-B) - sin(A-C)
= -sin(A-B) - sin(B-C) - sin(C-A)
= -sin(2A+C-180°) - sin(2B+A-180°) - sin(2C+B-180°) ###[since A+B+C=180°]
= sin(2A+C) + sin(2B+A) + (sin2C+B)
= L.H.S.

Using the last proof...

2) reduces to 4sinAsinBsinC/sinA + sinB + sinC

try from here..

5)cos2S + cos2(S-A) +cos2(S-B) +cos2(S-C) = 2+2cosAcosBcosC

LHS

= \frac{1}{2}[2cos^{2}S + 2cos^{2}(S-A) + 2cos^{2}(S-B) + 2cos^{2}(S-C)]

= \frac{1}{2}[4 + cos2S + cos(2S-2A) + cos(2S - 2B) + cos(2S-2C)]

= 2+ \frac{1}{2}[cos2S + cos(2S-2A) + cos(2S - 2B) + cos(2S-2C)]

= 2+ \frac{1}{2}[2cos(2S-A)cosA + 2cos(2S-B-C)cos(C-B)]

= 2+ \frac{1}{2}[2cos(C+B)cosA + 2cos(A)cos(C-B)]

= 2+ \frac{1}{2}[2cosA(cos(C+B) + cos(C-B))]

= 2+ \frac{1}{2}[2cosA(2cosCcosB)]

= \boxed{ 2+ 2cosAcosBcosC}

what is the proofs of following
A+B+C=180 degree.
Prove that,
1)sin(B+2C) + sin (C+2A) + sin(A+2B) = 4sin(B-C)/2* sin(C-A)/2 *sin(A-B)/2

2) (sin2A+sin2B+sin2C)/(sinA+sinB+sinC) = 8*sin(A/2)*sin(B/2)*sin(C/2)

3)sin(B+C-A) + sin(C+A-B) + sin(A+B-C) =4sinAsinBsinC