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If A+B+C=3pi/2. Then cos 2A +cos 2B+cos2...

If `A+B+C=3pi/2`. Then `cos 2A +cos 2B+cos2C` is equal to

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`

Text Solution

Verified by Experts

The correct Answer is:
D
`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`.
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