If A + B + C = 0, then prove that sin 2A + sin 2B + sin 2C = -4 sin A sin B sin C.
Answers
Answer:
sin 2A + sin 2B + sin 2C
= 4 sin A sin B sin C.
Step-by-step explanation:
Actually your question may be
A + B + C = 180⁰ prove that sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C.
Formula used:
1. sinC + sinD
= 2 sin ((C+D)/2) cos((C - D)/2)
2. cos(180⁰ - A) = - cosA
3. sin2A = 2 sinA cosA
sin 2A + sin 2B + sin 2C
= 2 sin((2A+2B)/2) cos((2A-2B)/2)
+ 2 sinC cosC
= 2 sin(A+B) cos(A-B) + 2 sinC cosC
= 2 sin(180⁰-C) cos(A-B) + 2 sinC cosC
= 2 sinC cos(A-B) + 2 sinC cosC
= 2 sinC [cos(A-B) + cosC]
= 2 sinC [cos(A-B) + cos(180⁰ -(A+B))]
= 2 sinC [cos(A-B) - cos(A+B)]
= 2 sinC [ 2 sinA sinB ]
= 4 sinA sinB sinC
I hope this answer helps you
Answer:
Step-by-step explanation:
A+B+C=0
A+B=-C -----1
LSH=sin2A+sin2B+sin2C
2sinA+BcosA-B+sin2C [USING IDENTITY sin2A]
putting A+B as -C from 1
2sin(-C)cosA-B+sin2C
-2sinCcosA-B+2sinCcosC
Taking common
2sinC[-cos(A-B)+cosC] [USING IDENTITY (cosC-cosD)]
2sinC[-2sin(C+A-B)/2sin(C-(A-B))/2]
putting again the value of C
-4sinC[sin(-A-B+A-B)/2sin(-A-B-A+B)/2]
-4sinC(sin-Bsin-A)
-4sinAsinBsinC
HENCE PROVED