PROVE SIN2A+SIN2B-SINC=4COSACOSBSINC

							
Dear bayana,
he double angle formula:
sin 2Θ = 2 sin Θ cos Θ

sin 2A + sin 2B - sin 2C
... = 2 sin A cos A + 2 sin B cos B - 2 sin C cos C 

Since A + B + C = π ;
A is a supplement angle of ( B + C )
B is a supplement angle of ( A + C )
C is a supplement angle of ( A + B )
TAKE NOTE that the sine of supplementary angles are equal !!!

sin 2A + sin 2B - sin 2C
... = 2 sin A cos A + 2 sin B cos B - 2 sin C cos C 
... = 2 sin ( B + C ) cos A + 2 sin ( A + C ) cos B - 2 sin ( A + B ) cos C

From the Sum of Angle Identity:
sin ( α + ß ) = sin α cos ß + cos α sin ß

sin 2A + sin 2B - sin 2C
... = 2 sin A cos A + 2 sin B cos B - 2 sin C cos C 
... = 2 sin ( B + C ) cos A + 2 sin ( A + C ) cos B - 2 sin ( A + B ) cos C
... = 2 ( sin B cos C + cos B sin C ) cos A 
..... ..... + 2 ( sin A cos C + cos A sin C ) cos B
..... ..... – 2 ( sin A cos B + cos A sin B ) cos C
... = 2 cos A sin B cos C + 2 cos A cos B sin C
..... ..... + 2 sin A cos B cos C + 2 cos A cos B sin C
..... ..... – 2 sin A cos B cos C – 2 cos A sin B cos C
... = 2 cos A cos B sin C + 2 cos A cos B sin C
... = 4 cos A cos B sin C
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first of all ur question is wrong
 
 
its      sin2a+sin2b-sin2c=4 cosa  cosb sinc
applying SIN2X=2SINXCOSX
A+B+C=180
AND                   A IS SUPPLEMENT ANGLE OF (B+C)
B IS SUPLEMET OF (A+C) AND C OF (A+B)
 
WE GET
SIN2A+SIN2B+SIN2C=2SIN(B+C)COSA+2SIN(A+C)COSB-2SIN(A+B)COSC
NOW APPLYING SUM PROPERTY
SIN(X+Y)=SINX COSY + COSX SINY
AND SOLVING WE GET                                    4COSACOBSINC