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If A+B+C =(3pi)/(2), then cos2A+cos2B+cos2C is equal to- |
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Answer» `1-4cosAcosBcosC` `=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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