Question

If A+B+C `=(3pi)/(2)`, then `cos2A+cos2B+cos2C` is equal to-
A. `1-4cosAcosBcosC`
B. `4sinAsinBsinC`
C. `1+2cosA+cosBcosC`
D. `1-4sinAsinBsinC`

Answer 1

Correct Answer - d
`cos2A+cos2B+cos2C=2cos(A+B)cos(A-B)+cos2C`
`=2cos((3pi)/2-C) cos(A-B)+cos2C therefore A+B+C=(3pi)/(2)`
`=-2sinCcos(A-B)+1-2sin^(2)C=1-2sinC[cos(A-B)+sinC]`
`=1-2sinC[cos(A-B)+sin((3pi)/(2)-(A+B))]`
`=1-2sinC[cos(A-B)-cos(A+B)]=1-4sinBsinC`
`=1-2sinC[cos(A-B)+sin((3pi)/2-(A+B))]`
`=1-2sinC[cos(A-B)-cos(A+B)]=1-4sinAsinBsinC`