Pre-Calculus

If sinx=-1/2 and x terminates in the third quadrant, find the exact value of tan2x.

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asked by Brenda
  1. tan ( 2 x ) = 2 ∙ tan x / ( 1 - tan² x )

    tan x = sin x / cos x

    cos x = ± √ ( 1 - sin² x )

    In quadrant III cosine is positive so:

    cos x = √ ( 1 - sin² x )

    cos x = √ [ 1 - ( - 1 / 2 )² ]

    cos x = √ ( 1 - 1 / 4 )

    cos x = √ ( 4 / 4 - 1 / 4 )

    cos x = √ ( 3 / 4 )

    cos x = √3 / √4

    cos x = √3 / 2


    tan x = sin x / cos x

    tan x = ( - 1 / 2 ) / ( √3 / 2 ) = ( - 1 ∙ 2 ) / ( √3 ∙ 2 ) = - 1 / √3


    tan x = - 1 / √3


    tan ( 2 x ) = 2 ∙ tan x / ( 1 - tan² x )

    tan ( 2 x ) = 2 ∙ ( - 1 / √3 ) / [ 1 - ( - 1 / √3 )² ] =

    - 2 √3 ) / [ 1 - ( 1 / 3 ) ] =

    - 2 √3 ) / ( 3 / 3 - 1 / 3 ) =

    ( - 2 / √3 ) / ( 2 / 3 ) =

    ( - 2 ∙ 3 ) / ( 2 ∙ √3 ) =

    - 3 / √3 =

    - √3 ∙ √3 / √3 = - √3


    tan ( 2 x ) = - √3


    By the way:

    x = 11 π / 6 = 330°

    2 x = 11 π / 3 = 660°

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    posted by Bosnian
  2. sin π/6 = 1/2, so that is your reference angle

    In QIII, x = π+π/6 = 7π/6
    so, 2x = 7π/3 = 2π+π/3, in QI
    So, tan2x = √3

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    posted by Steve

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