I'm going to guess this is supposed to read: Solve for # tan theta # in
#{ tan theta + tan 3 theta} /{1 - tan theta tan 3 theta } = -1#
We recognize this as the tangent sum angle formula:
#tan(a+b) = {tan a + tan b}/{1 - tan a tan b } #
So our equation becomes,
#tan 4theta= -1#
It's tempting to try to jump right to the answer, but a prefer a standard approach that uses the fact that #cos x = cos a# has solutions #x=\pm a + 360^circ k quad # for integer #k#.
# {sin 4 theta}/ {cos 4 theta} = -1#
#-sin 4theta = cos 4 theta#
#sin(-4theta) = cos 4 theta #
#cos (90^circ + 4theta) = cos 4 theta #
# 90^circ + 4theta = \pm 4 theta + 360^circ k #
Only the minus sign gets us solutions:
# 8 theta = -90^circ + 360^circ k#
#theta = - 11.25^circ + 45^circ k #
EDIT: Oops, I solved for #theta# when I should have solved for #tan theta.#
DId you ever notice there's no half angle formula for tangent in terms of tangent? (There is one we can get with the quadratic formula.) The typical path is through the sine and cosine.
# tan(a/2) = {1 - cos a}/sin a#
This part is tricky because there are two values for the pair #(cos a, sin a)# that give any given tangent. In one the cosine and the sine are negated relative to the other.
# cos(a/2) = pm sqrt{1/2( 1 + cos a)}#
#sin(a/2) = pm sqrt{1/2(1-cos a)}
#
We have eight #theta#s that will give rise to four unique values of #tan theta#. Let's calculate them.
#4 theta = -45^circ or 145^circ#
#cos(-45^circ)=1/sqrt 2 quad quad quad quad sin(-45^circ) = -1/sqrt{2}#
#cos(-45^circ/2) = \sqrt{1/2 (1 + sqrt{2}/2) }#
#cos(-22.5^circ) =1/2 sqrt{2 + sqrt{2}} #
# sin(-22.5^circ) = - 1/2 sqrt{2 -sqrt{2}}#
# tan(-11.25 ^circ) = { 1 - cos(-22.5^circ)} / {sin (-22.5^circ) } = { 1 - 1/2 sqrt{2 + sqrt{2}} } /{- 1/2 sqrt{2 -sqrt{2}} } ##= 1 + sqrt(2) - sqrt(2 (2 + sqrt(2))) #
# tan (a+ 45^circ) = {tan a + tan 45^circ}/{1 - tan a tan 45^circ } = {1 + tan a}/{1 -tan a } #
# tan(-11.25^circ + 45^circ) = tan(33.75^circ) ={1 + ( 1 + sqrt(2) - sqrt(2 (2 + sqrt(2))) )}/{1 - ( 1 + sqrt(2) - sqrt(2 (2 + sqrt(2))) ) } = 1 - sqrt(2) + sqrt(4 - 2 sqrt(2))
#
# tan(-11.25^circ + 90^circ) = 1/tan(-11.25^circ/2) = ##= 1 + sqrt(2) + sqrt(2 (2 + sqrt(2))) #
# tan (a- 45^circ) = {tan a - 1 }/{ tan a + 1 } #
# tan(-11.25^circ - 45^circ) = tan(-56.25^circ) = { (1 + sqrt(2) - sqrt(2 (2 + sqrt(2))) ) - 1 }/{ ( 1 + sqrt(2) - sqrt(2 (2 + sqrt(2))) ) + 1 } ## = 1 - sqrt(2) (1 + sqrt(2 - sqrt(2)))
#
