If #sin s = -12/13# and #sin t = 4/5#, what is #cos(s+t)# and #cos (s-t)#?

We will need the values of #cos(s) and cos(t)#, therefore, we shall use the identity:

#cos(x) = +-sqrt(1-sin^2(x))" [1]"#

Substitute #sin(s)= -12/13# into equation [1]:

#cos(s) = +-sqrt(1-(-12/13)^2)#

#cos(s) = +-sqrt(169/169-144/169)#

#cos(s) = +-sqrt(25/169)#

#cos(s) = +-5/13" [2]"#

We are told that #s# is in the fourth quadrant, therefore, we shall choose the positive value:

#cos(s) = 5/13#

Substitute #sin(t)= 4/5# into equation [1]:

#cos(t) = +-sqrt(1-(4/5)^2)#

#cos(t) = +-sqrt(25/25-16/25)#

#cos(t) = +-sqrt(9/25)#

#cos(t) = +-3/5#

We are told that #t# is in the second quadrant, therefore, we shall choose the negative value:

#cos(t) = -3/5" [3]"#

Using the identity,

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

, we substitute #cos(s) = 5/13, cos(t) = -3/5, sin(s) = -12/13, and sin(t) = 4/5#:

#cos(s+t) = (5/13)(-3/5) - (-12/13)(4/5)#

#cos(s+t) = 33/65#

Using the identity,

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

, we substitute #cos(s) = 5/13, cos(t) = -3/5, sin(s) = -12/13, and sin(t) = 4/5#:

#cos(s-t) = (5/13)(-3/5) + (-12/13)(4/5)#

#cos(s-t) = -63/65#