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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