# trigonometry

Given tanx=8/15 when pi < x < 3pi/2, find tan x/2.

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1. Use tan (2A) = 2tanA/(1 - tan^2 A)
or tanx = 2tan (x/2)(1 - tan^2 (x/2)) , let tan x/2 = y for easier typing
8/15 = 2y/(1 - y^2)
30y = 8 - 8y^2
8y^2 + 30y - 8 = 0
4y^2 + 15y - 4 = 0
(4y - 1)(y + 4) = 0
y = 1/4 or y = -4
tan x/2 = 1/4 or tan x/2 = -4
but x was in quadrant III , so x/2 must be in quadrant II
and in II, the tangent is negative,
so tan x/2 = -4

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2. if tanx = 8/15 in QIII then
sinx = -8/√161
cosx = -15/√161
so
tanx/2 = (1-cosx)/sinx = (1 + 8/√161)/(-15/√161) = -(8+√161)/15
makes sense, since x/2 will be in QII where tangent is negative

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3. oops - go with Reiny.
Do you see my mistake?

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4. Tanx = -8/-15 = 8/15.
X = 28.1o S. of W. (Q3). = 208.1o CCW.
x/2 = 208.1/2 = 104o CCW = 76o N. of W.
Tan(x/2) = Tan104 = -4.0.

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