Question

Question #e1771

Answer 1

`tan 165 = sqrt3 - 2`

Explanation:

Call tan 165 = tan t --> tan 2t = tan 330
Trig table and unit circle give:.
`tan 330 = tan (-30 + 360) = tan (-30) = - tan 30 = - 1/sqrt3`
Use trig identity:
`tan 2t = (2tan t)/(1 - tan^2 t)`
In this case :
`-1/sqrt3 = (2tan t)/(1 - tan^2 t)`
`tan^2 t - 1 = 2sqrt3tan t`
`tan^2 t - 2sqrt3tan t - 1 = 0`
Solve this quadratic equation for tan t using improved quadratic formula (Socratic Search):
`D = d^2 = b^2 - 4ac = 12 + 4 = 16` --> `d = +- 4`.
There are 2 real roots:
#tan t = - b/(2a) +- d/(2a) = sqrt3 +- 2
tan t = sqrt3 + 2, and tan t = sqrt3 - 2
Since (165) is in Quadrant II, tan 165 is negative, then take the negative value.
tan (165) = tan t = sqrt3 - 2