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

Options

A. `1-4cos A cos B cos C`
B. `4 sin A sin B sinC`
C. `1+2cos A cos B cos C`
D. `1-4sin A sin B sinC`

Answer

D

Explanation

`cos 2A+cos2B+cos 2C`
`=2cos(A+B)cos(A-B)+cos2C`
`=2cos(A+B)cos(A-B)+cos2C`
`=2cos((3pi)/(2)-C)cos(A-B)+cos2C`
`=2cos((3pi)/(2)-C)cos(A-B)+cos2C`
`=-2sin C cos (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)]`
`=1-4sinA sin B sin C`.