Prove that cos3 Acos 3A + sin3A sin 3A= cos32A
Dear Student,
LHS = cos3A cos3A + sin3A sin3A
= cos3A (4cos3A - 3cosA) + sin3A (3sinA - 4sin3A) (using identities of cos3A and sin3A)
= 4cos6A - 3cos4A + 3sin4A - 4sin6A
= 4 (cos6A - sin6A) + 3 (sin4A - cos4A)
= 4 ( (cos2A)3 - (sin2A)3) + 3 ( (sin2A)2 - (cos2A)2 )
= 4 [ (cos2A - sin2A) (cos4A + cos2Asin2A + sin4A)] + 3 [(sin2A+cos2A)(sin2A - cos2A)]
using a3 - b3 = (a-b)(a2+ab+b2) and a2-b2=(a-b)(a+b)
= 4 [ cos2A ((sin2A+cos2A)2 - cos2Asin2A) ] + 3 [ 1 (-cos2A)]
= 4 [cos2A (1 - cos2Asin2A )] - 3 cos2A
= 4cos2A - 4cos2A cos2Asin2A - 3cos2A
= cos2A - 4cos2A cos2Asin2A
= cos2A (1 - 4cos2Asin2A )
= cos2A (1 - sin22A) (Since 2sinA cosA = sin2A )
= cos2A cos22A
= cos32A
= RHS
Hence proved
Regards
LHS = cos3A cos3A + sin3A sin3A
= cos3A (4cos3A - 3cosA) + sin3A (3sinA - 4sin3A) (using identities of cos3A and sin3A)
= 4cos6A - 3cos4A + 3sin4A - 4sin6A
= 4 (cos6A - sin6A) + 3 (sin4A - cos4A)
= 4 ( (cos2A)3 - (sin2A)3) + 3 ( (sin2A)2 - (cos2A)2 )
= 4 [ (cos2A - sin2A) (cos4A + cos2Asin2A + sin4A)] + 3 [(sin2A+cos2A)(sin2A - cos2A)]
using a3 - b3 = (a-b)(a2+ab+b2) and a2-b2=(a-b)(a+b)
= 4 [ cos2A ((sin2A+cos2A)2 - cos2Asin2A) ] + 3 [ 1 (-cos2A)]
= 4 [cos2A (1 - cos2Asin2A )] - 3 cos2A
= 4cos2A - 4cos2A cos2Asin2A - 3cos2A
= cos2A - 4cos2A cos2Asin2A
= cos2A (1 - 4cos2Asin2A )
= cos2A (1 - sin22A) (Since 2sinA cosA = sin2A )
= cos2A cos22A
= cos32A
= RHS
Hence proved
Regards