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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

4sinAsinBsinC

B

4cosAcosBcosC

C

1-4sinAsinBsinC

D

1-4cosAcosBcosC

Text Solution

Verified by Experts

The correct Answer is:
C
`cos2A+cos2B+cos2C`
`=2cos(A+B)cos(A-B)+1-2sin^(2)C`
`=2cos((3pi)/(2)-C)cos(A-B)+1-2sin^(2)C" "[becauseA+B+C=270^(@)impliesB+A=(3pi)/(2)-C]`
`=1-2sinC[cos(A-B)+sinC]`
`=1-2sinC[cos(A-B)-cos(A+B)]`
`=1-4sinAsinBsinC`
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MHTCET PREVIOUS YEAR PAPERS AND PRACTICE PAPERS-FACTORIZATION FORMULAE -EXERCISE 2
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