How do you use a half-angle formula to simplify tan 105?

2 Answers
Jun 11, 2015

A derivation for tan(x/2) is at the bottom.

tan(x/2) = pmsqrt((1-cosx)/(1+cosx))

depending on the quadrant of 105^o (quadrant II, thus the horizontal axis < 0 on the unit circle).

tan 105^o = tan ((7pi)/12) = tan (1/2*(7pi)/6) = -sqrt((1-cos((7pi)/6))/(1+cos((7pi)/6)))

= -sqrt((1+sqrt3/2)/(1-sqrt3/2))

= -sqrt(((2+sqrt3)/2)/((2-sqrt3)/2))

= -sqrt((2+sqrt3)/(2-sqrt3))

= -sqrt((2+sqrt3)^2/((2-sqrt3)(2+sqrt3)))

= -sqrt((2+sqrt3)^2)/(1)

And since we already specified the quadrant, there's no need for pm (and 2 + sqrt3 > 0 of course).

= -(2 + sqrt3)

= -2 - sqrt3


You can derive the half-angle formula if you don't remember it.
sin^2(x) = (1-cos(2x))/2

Similarly:

sin^2(x/2) = (1-cos(x))/2

Thus:

|sin(x/2)| = sqrt((1-cosx)/2)

Similarly:

cos^2(x) = (1+cos(2x))/2

cos^2(x/2) = (1+cosx)/2

|cos(x/2)| = sqrt((1+cosx)/2)

Now we can divide them.

|sin(x/2)/(cos(x/2))| = |tan(x/2)| = sqrt((1-cosx)/2) / sqrt((1+cosx)/2)

= sqrt((1-cosx)/(1+cosx))

Jun 12, 2015

Simplify tan 105
Use trig identity: tan (a + b) = (tan a + tan b)/(1 - tan a. tan b)

Explanation:

tan 105 = tan (45 + 60)

tan 45 = 1 ; tan 60 = sqrt3

tan 105 = (1 + sqrt3)/(1 - sqrt3)