Question

If a+b+c=180 then prove that cos2a+cos2b-cos2c=1-4sinasinbcosc

Answer 1

cos2A + cos2B - cos2C = cos2A - 2sin[( 2B + 2C)/2]sin[(2B - 2C)/2] = cos2A - 2sin( B + C )sin(B - C) We have : A + B + C = 180° => B + C = 180° - A = cos2A - 2sin( 180° - A )sin(B - C) = cos2A - 2( sin180°cosA - sinAcos180° )sin( B - C ) = cos2A + 2sinAsin(B - C) = 1 - 2sin²A + 2sinAsin(B - C) = 1 - 2sinA[ sinA - sin(B - C) ] = 1 - 2sinA( 2cos[(A + B - C)/2]sin[(A - B + C)/2] ) = 1 - 4sinA.cos( A + B - C )/2.sin(A - B + C ) /2 We have : A + B + C = 180° => A + B = 180° - C And : A + C = 180° - B = 1 - 4sinA.cos[( 180° - C - C )/2].sin[( 180° - B - B )/2] = 1 - 4sinA.cos( -2C )/2.sin( -2B)/2 = 1 - 4sinA.cos(-C).sin(-B) = 1 - 4sinA.sinB.cosC