If A + B + C = π. Then
Column AColumn B1.sin2A + sin2B + sin2CA.cosA2 cos B2 cos C22.sinA + sinB + sinCB.1+4sinA2 sinB2 sinC23.cos2A + cos2B + cos2CC.4sinA sinB sinC4.cosA + cosB + cosCD.-1-4cosA + cosB + cosC
1 - C, 2 - A, 3 - D, 4 - B
1. sin2A + sin2B + sin2C
we need to just simplify the given expression.
We know,
sinA + sinB = 2 sin(A+B)2. cos(A−B)2
2sin(2A+2B)2.cos(2A+2B)2 + sin 2[π - (A+B)]
2 sin(A+B).cos(A-B)+ sin (2π - 2(A+B))
2 sin(A+B).cos(A-B) - sin 2(A+B)
2 sin(A+B).cos(A-B) - 2 sin(A+B). cos(A+B) {sin 2A = 2 sinA cosA}
2 sin(A+B){cos(A-B) - cos(A+B)}
= 2 sin(π - C) [2sin(A−B+A+B)2sin(A+B−A+B)2]
= 2 sinC{2 . sinA . sinB}
= 4 sinA sinB sinC
2.sinA sinB sinC
2 sin(A+B)2 . cos(A−B)2 + sin (π - (A+B)) {C = π - (A+B)}
= 2 sin(A+B)2 . cos(A−B)2 + sin(A+B)
= 2 sin(A+B)2 . cos(A−B)2 + 2 sin(A+B)2 . cos(A+B)2 [sinA=2sinA2.cosA2]
= 2 sin(A+B)2 [cosA−B2+cosA+B2]
= 2sin(π−C2)×[2cos(A−B+A+B2)2×cos(A+B−A+B2)2]
= 2 sin (π2−C2).2 cosA2 . cosB2
= 2 cosC2 . 2cosA2 . cosB2
= 4 cosA2 cosB2 cosC2
3. cos2A + cos2B + cos2C
2 cos (2A+2B)2 cos(2A−2B)2 + cos 2(π - (A+B))
= 2 cos(A+B). cos(A-B) + {cos (2π - 2(A+B))}
= 2 cos(A+B). cos(A-B) + cos 2(A+B)
= 2 cos(A+B). cos(A-B) + 2cos2(A+B) - 1 {cos 2A = 2cos2 A - 1}
= 2 cos(A+B) {cos(A-B) + cos(A+B)} - 1
= 2cos(π−C){2cos(A−B+A+B)2cos(A+B−A+B)2}−1
= -2cos C {2 cosA cosB} - 1
= -4 cosA cosB cosC - 1
4. cosA + cosB + cosC
= 2 cos (A+B)2 . cos (A−B)2 + cos{π - (A+B)}
= 2 cos (A+B)2 . cos (A−B)2 +{-cos (A+B)}
= 2 cos (A+B)2 . cos (A−B)2 - cos(A+B)
= 2 cos (A+B)2 . cos (A−B)2 - [2cos2(A+B)2−1] {cos 2A = 2 cos2 A - 1}
= 2 cos (A+B)2 . cos (A−B)2 - 2cos2(A+B)2+1
= 2 cos (A+B)2 [cos(A−B)2−cos(A+B)2] + 1
= 2 cos (π+C)2 [2sinA2.sinB2+1]
= 2 sinC2 . 2 sinA2 . sinB2 + 1
= 1 + 4sinA2 sinB2 sinC2