# Pre-Calculus

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

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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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