Question

cos²A+cos²B+2cosA×cosB×cosC=sin²C​

Answer 1

Answer:

2 sinA . sinB . cosC

= sinA (2 sinB.cosC)

= sinA (sin(180-A) + sin(B-C) )

= sinA (sinA + sin(B-C) )

= sin²A + sinA sin(B-C)

= sin²A + {2sinA sin(B-C) / 2}

= sin²A + {cos(A-B+C) - cos(A+B-C) / 2}

= sin²A + {cos(A+C-B) - cos(A+B-C) / 2}

= sin²A + {cos(180-B-B) - cos(180-C-C) / 2}

= sin²A + {cos(180-2B) - cos(180-2C) / 2}

= sin²A - {cos2B + cos2C / 2}

= ( 2sin²A - cos2B + cos2C ) / 2

= 1/2 { 2sin²A - cos2B + cos2C }

= 1/2 { 2sin²A - 1+2sin²B + 1-2sin²C }

=

1/2 { 2sin²A + 2sin²B - 2sin²C }

= sin²A + sin²B - sin²C

= sin²A + sin²B - sin²C = 2 sinA . sinB . cosC

= 1 - cos²A + 1-cos²B -1+cos²C =2sinA.sinB.cosC

= cos²A + cos²B - cos²C

= 1- 2sinA.sinB.cosC

Hence Proved