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[Post New]5 Apr 2008 22:06:19 IST
    Subject: Re:graph
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srini (305)

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Olaaa!! Perrrfect answer. 49  [79 rates]
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Hey,
in [0.pi/2], sin^-1(sinx)=x =>\cosx\=x^2;
    [pi/2,pi]       pi/2 <x<pi
                       =>pi-x   => |cosx|=(pi-x)^2
 
Its this way?
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