Posted by briane on Monday, April 7, 2014 at 3:12pm.
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

math  Steve, Monday, April 7, 2014 at 3:29pm
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) = (1cos2x)/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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