cos2A+cos2B-cos2C=1-4sinAsinBsinC

Asked by irenegladis | 5th Apr, 2009, 09:08: AM

Expert Answer:

we know that cos u - cos v =-2 sin(u+v)/2  sin(u- v)/2

cos2A-cos2C = -2sin(A+C)  sin(A-C) = -2 sinB sin(A-C)

and cos2B = 1 - 2sin2B

cos2A+cos2B-cos2C= -2 sinB sin(A-C) +1 - 2sin2B

=1 - 2sinB (sin(A-C) + sinB )

=1- 2 sinB 2sin(A-C+B) /2   cos(A-C-B)/2

=1 - 4 sinB sin(π-C-C) /2   cos(A-(π-A))/2

=1- 4 sinB cosC sinA

so instead of sinC, cosC will be there.

Answered by  | 9th Apr, 2009, 02:34: AM

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