if tan 2x= -b/a, pi/2 <=2x<=pi, then determine an expression for sinx cosx in terms of a and b.
*please show the calculation

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  1. the two legs are a and b, and the hypotenuse is √(a^2+b^2). So, assuming a,b > 0,

    So, in QII,
    sin2x = b/√(a^2+b^2)
    cos2x = -a/√(a^2+b^2)

    sin^2(x) = (1-cos2x)/2 = (1+a/√(a^2+b^2))/2
    = (a+√(a^2+b^2)) / 2√(a^2+b^2)

    and similarly for cosx

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