If sin2x=7/25, siny=-2/sq root of 13, cot y>0, and -pi/4<x<pi/4, find sin(2y-a)

Answer 1

If siny = -2/√13 , and coty ≥0, then y is in quad III
so cosy = -3/√13
then sin 2y = 2sinycosy
= 2(-2/√13)(-3/√13) = 12/13
cos 2y = cos^2 y - sin^2 y
= 9/13 - 4/13 = 5/13

so sin 2y = 12/13 and cos 2y = 5/13

if sin 2x = 7/25

remember cos^2 (2x) = 1 - sin^2 (2x)
= 1 - 49/625 = 576/625
then cos 2x = 24/25 , (or from right-angled triangle 7-24-25)

and cos (2x) = 1 - 2sin^2 (x)
2sin^2 x = 1 - cos2x
2sin^2 x = 1 - 576/625 = 49/625
sin^2 x = 49/1250
sin x = ±7/25√2 --->(x = 11.4° for -45°<x<45° )
sinx = 7/ 25√2
and cos^2 x = 1 - sin^2 x = 1201/1250
cos x = ± √1201/ 25√2
cosx = √1201/ 25√2

so sinx = 7/ 25√2 , cosx = √1201/ 25√2

now finally,

arhhh, you have a typo

I am sure you meant:
sin(2y - x)
= sin 2y cosx - cos 2y sinx
= (plug in my values from above)

= (12/13)(√1201/25√2) - (5/13)(7/25√2)
= (12√1201 - 35)/(325√2)

What a mess! I did not write this out on paper first, so you better check my arithmetic

Answer 2

I checked with my calculator and actually found the angles.
It works !!!