How do you use the half angle formulas to determine the exact values of sine, cosine, and tangent of the angle #165^circ#?

1 Answer
Jan 5, 2018

#sin(165^circ) = sqrt(2-sqrt(3))/2#,

#cos(165^circ) = - sqrt(2+sqrt(3))/2#,

#tan(165^circ)= -sqrt(7-4sqrt(3))#

Explanation:

#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))#