posted by Matt .
Given that sinx = 7 /25 pi/2 <or equal to x <or equal to pi Find the value of cos 2x
sinx = 7/25 , π/2 ≤ x ≤ π
so x is in quadrant II , making 2x fall in quadrant IV
So the opposite side is 7, and the hypotenuse is 25, so the adjacent side would be 24 by Pythagoras ,
and cosx = -24/25
cos 2x = cos^2 x - sin^2 x
= 576/625 - 49/625 = 527/625
Just a note:
If π/2 ≤ x ≤ π, π ≤ 2x ≤ 2π which means 2x could be in QIII or QIV.
In this case, since x > 3π/4, 2x>3π/2, and is indeed in QIV, but it need not have been so.