# math

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

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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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