Let sin s 13 with s in quadrant ii and let cos t 34 with t

Let sin s = 1/3, with s in quadrant ii and let cos t =3/4, with t in quadaratn i. Find each of the following:

a) cos (s+t)

b) tan (s-t)
c) cos 2s

d) sin t/2

Solution

sin s = 1/3, cos t = 3/4

cos s = +- ( 1- sin² s)^1/2

= +- ( 1- ( 1/3)²)^1/2

cos s = - 8^1/2 / 3 ( s - II quadrant, in II quad. cos s = -ve)

sin t = +- ( 1- cos² t)^1/2

= + - ( 1- 9/16) ^1/2

= + 7^1/2 / 4 ( sin t = +ve in I quadrant)

cos (s+t) = cos s*cos t - sin s*sin t

= ( - 8^1/2 /3)* ( 3/4) - (1/3)* (7^1/2 /4)

= - 0.927

-----------------------------------------------------------------------

tan( s- t) = sin ( s-t) / cos ( s-t)

= sin s* cos t - cos s*sin t / cos s *cos t + sin s *sint

on putting values = - 1.79

-----------------------------------------------------------------

cos 2s = 2*cos²s - 1

= 2* ( - 8^1/2 / 3) - 1

= 0.777

---------------------------------

sin t/2 = ?

from cos t = 1-2*sin² t/2

sin² t/2 = 1- cos t/2

= 1- (3/4) / 2

= 1/8

sin t/2 = 1/8^1/2   ( since t is in I quadrant)