#165^circ# is in QII, so we know that #sin(165^circ)# is positive and #cos(165^circ)# is negative.
#165^circ = (330^circ)/2#, so we'll use #330^circ# in the half angle formulas. #cos(330^circ) = -sqrt(3)/2#.
#sin(x/2) = pm sqrt((1-cos(x))/2)#
#cos(x/2) = pm sqrt((1+cos(x))/2)#
Combining what we know so far we can find sine:
#sin(165^circ) = sin(330^circ/2) = sqrt((1-cos(330^circ))/2)#
#=sqrt((1-(-sqrt(3)/2))/2) = sqrt(((2-sqrt(3))/2)/2)#
#=sqrt((2-sqrt(3))/4) = sqrt(2-sqrt(3))/2#
Now let's find cosine in much the same way:
#cos(165^circ) = cos(330^circ/2) = -sqrt((1+cos(330^circ))/2)#
#=-sqrt((1+(-sqrt(3)/2))/2) = -sqrt(((2+sqrt(3))/2)/2)#
#=-sqrt((2+sqrt(3))/4) = -sqrt(2+sqrt(3))/2#
So we've found:
#sin(165^circ) = sqrt(2-sqrt(3))/2#
and
#cos(165^circ) = - sqrt(2+sqrt(3))/2#
To find #tan(165^circ)# we'll use the identity:
#tan(x) = sin(x)/cos(x)#.
So,
#tan(165^circ) = sin(165^circ)/cos(165^circ)#
#=(sqrt(2-sqrt(3))/2)/(- sqrt(2+sqrt(3))/2) = -sqrt(2-sqrt(3))/(sqrt(2+sqrt(3)) #
we can do more work on this, if we want, although that's finished answer for many people:
#-sqrt(2-sqrt(3))/(sqrt(2+sqrt(3)))= -sqrt((2-sqrt(3))/(2+sqrt(3))) #
#=-sqrt((2-sqrt(3))/(2+sqrt(3))* (2-sqrt(3))/(2-sqrt(3))) #
#=-sqrt((7-4sqrt(3))/(1)) = -sqrt(7-4sqrt(3))#
