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If A+B+C=(3pi)/2, t h e ncos2A+cos2B+cos...

If `A+B+C=(3pi)/2, t h e ncos2A+cos2B+cos2C` is equal to `1-4cos Acos BcosC` `4sinAsinBsinC` `1+2cosAcosBcosC` `1-4sinAsinBsinC`

A

`1-4 cosA cos B cosC`

B

`4 sinA sin B sinC`

C

`1+2cos cos B cosC`

D

`1-4sinA sin B sinC`

Text Solution

Verified by Experts

The correct Answer is:
D
`cos 2A+cos2B+cos2C=2cos(A+B)cos(A-B)+cos 2C`
`=2 cos((3pi)/(2)-C)cos(A-B)+cos2C because A+B+C=(3pi)/(2)`
`=-2sin C cos(A-B)+1-2sin^(2)C=1-2sin ^(2)C=1-2sinC[cos(A-B)+sinC)`
`=1-2sinC[cos(A-B)+ sin((3pi)/(2)-( A+B))]`
`=1-2sin C[cos(A-B)-cos(A+B)]=1-4 sin A sin B sin C`
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