if sin(x)= 1/3 and sec(y)=13/12,where x and y lie between 0 and pi/2, evaluate the expression using trigonometric identities. sin(2y)

Answer 1

sinx = 1/3

secy = 13/12

cosy =1/secy = 12/13

sinx = (sqr(13^2 -12^2)/13

= sqr(169-144)/13

= sqr(25)/13 = 5/13

sin2y = 2sinycosy

= 2(5/13)(12/13)

= 120/169

= about 0.71

whatever answer you come up with has to be between -1 and +1 for sine of any angle. 0.71 is in the right range

2y = sin^-1(120/169)

2y = about 45. 24

y = about 45.24/2 = 22.6 degrees

sinx = 1/3

x = sin^-1(1/3

x = about 19. 5 degrees

they both seem to be relatively close, a coincidence? or is there more to this problem?