Given: #sin(theta) = 2/3# and #cos(varphi) = 1/4#
The identity for #cos(theta-varphi)# is:
#cos(theta-varphi) = cos(theta)cos(varphi)+sin(theta)sin(varphi)#
Please observe that we must compute obtain the values of #cos(theta)# and #sin(varphi)#.
Use the identity:
#cos(theta) = +-sqrt(1-sin^2(theta))#
Substitute #sin^2(theta) = (2/3)^2#:
#cos(theta) = +-sqrt(1-(2/3)^2)#
#cos(theta) = +-sqrt(9/9-4/9)#
#cos(theta) = +-sqrt(5/9)#
#cos(theta) = +-sqrt5/3#
We are told that #theta# is acute, therefore, we choose the positive alternative:
#cos(theta) = sqrt5/3#
Use the identity:
#sin(varphi) = +-sqrt(1-cos^2(varphi))#
Substitute #cos^2(varphi) = (1/4)^2#:
#sin(varphi) = +-sqrt(1-(1/4)^2)#
#sin(varphi) = +-sqrt(16/16-1/16)#
#sin(varphi) = +-sqrt(15/16)#
#sin(varphi) = +-sqrt15/4#
We are told that #varphi# is acute, therefore, we choose the positive alternative:
#sin(varphi) = sqrt15/4#
We, now, have all 4 values that we need to the identity:
#cos(theta-varphi) = cos(theta)cos(varphi)+sin(theta)sin(varphi)#
Substitute the respective values:
#cos(theta-varphi) = (sqrt5/3)(1/4)+(2/3)(sqrt15/4)#
#cos(theta-varphi) = (sqrt5+2sqrt15)/12#
