Question

cos2a+cos2b-cos2c=1-4sinasinbcosc

Answer 1

s2A + 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 

2) cos2A + cos2B - cos2C 
If A = 75° and B = 15° => C = 180° - 75° - 15° = 90° 
We know : cos2A + cos2B - cos2C = 1 - 4sinA.sinB.cosC 
= 1 - 4.sin75°.sin15°.cos90° 

= 1 - 0 

= 1 

3) cos2A + cos2B + cos2C 

= 2cos[(2A + 2B)/2].cos[(2A - 2B)/2] + cos2C 

= 2cos(A + B).cos(A - B) + cos2C 

We have : A + B + C = 180° => A + B = 180° - C 

= 2cos( 180° - C ).cos(A - B) + cos2C 

= 2( cos180°cosC + sin180°sinC ).cos(A - B) + cos2C 

= 2( -cosC ).cos(A - B) + 2cos²C - 1 

= -2cosC.cos(A - B) + 2cos²C - 1 

= 2cosC[ cosC - cos(A - B)] - 1 

= 2cosC( -2sin[(C + A - B)/2]sin[(C - A + B)/2] ) - 1 

= -4cosC.sin[(C + A - B)/2]sin[(C - A + B)/2] - 1 

We have : C + A = 180° - B and C + B = 180° - A 

= -4cosC.sin[(180° - B - B)/2]sin[(180° - A - A)/2] - 1 

= -4cosC.sin( 90° - B).sin(90° - A ) - 1 

= -4cosC.cosB.cosC - 1 

i) sin2A + sin2B + sin2C 

= 2sin[(2A + 2B)/2].cos[( 2A - 2B)/2] + sin2C 

= 2sin(A + B).cos(A - B) + sin2C 

We have : A + B + C = 180° => A + B = 180° - C 

= 2sin( 180° - C).cos(A - B) + sin2C 

= 2( sin180°cosC - cos180°sinC ).cos(A - B) + sin2C 

= 2sinC.cos(A - B) + 2sinCcosC 

= 2sinC[ cos(A - B) + cosC ] 

= 2sinC( 2cos[(A - B + C)/2].cos[(A - B - C)/2] ) 

= 4sinC.cos[(A - B + C)/2].cos[(-A + B + C)/2] 

We have : A + C = 180° - B ; B + C = 180° - A 

= 4sinC.cos[( 180° - B - B)/2].cos[( 180° - A - A )/2] 

= 4sinC.cos( 90° - B).cos( 90° - A ) 

= 4sinC.sinB.sinA