Question #e1771

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Question #e1771

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Answer 1 · 2017-02-13T14:18:59

$tan 165 = \sqrt{3} - 2$ $tan 165 = sqrt3 - 2$

Explanation:

Call tan 165 = tan t --> tan 2t = tan 330
Trig table and unit circle give:.
$tan 330 = tan (- 30 + 360) = tan (- 30) = - tan 30 = - \frac{1}{\sqrt{3}}$ $tan 330 = tan (-30 + 360) = tan (-30) = - tan 30 = - 1/sqrt3$
Use trig identity:
$tan 2 t = \frac{2 tan t}{1 - {tan}^{2} t}$ $tan 2t = (2tan t)/(1 - tan^2 t)$
In this case :
$- \frac{1}{\sqrt{3}} = \frac{2 tan t}{1 - {tan}^{2} t}$ $-1/sqrt3 = (2tan t)/(1 - tan^2 t)$
${tan}^{2} t - 1 = 2 \sqrt{3} tan t$ $tan^2 t - 1 = 2sqrt3tan t$
${tan}^{2} t - 2 \sqrt{3} tan t - 1 = 0$ $tan^2 t - 2sqrt3tan t - 1 = 0$
Solve this quadratic equation for tan t using improved quadratic formula (Socratic Search):
$D = d^{2} = b^{2} - 4 a c = 12 + 4 = 16$ $D = d^2 = b^2 - 4ac = 12 + 4 = 16$ --> $d = \pm 4$ $d = +- 4$ .
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

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Question #e1771

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