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If A+B+C=(3pi)/2 , then cos2A+cos2B+cos2...

If `A+B+C=(3pi)/2` , then `cos2A+cos2B+cos2C` is equal to `1-4cos A\ cos B\ cos C` b. `4sinA\ s in B\ s in C` c.`1+2cos A\ cos B\ cos C` d. `1-4s in A\ s in BsinC`

A

`1-4cosAcosBcosC`

B

`4sinAsinBsinC`

C

`1+2cosA+cosBcosC`

D

`1-4sinAsinBsinC`

Text Solution

Verified by Experts

The correct Answer is:
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`
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