If sinx=-1/2 and x terminates in the third quadrant, find the exact value of tan2x.
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
To find the exact value of tan(2x), we need to use trigonometric identities and properties.
First, let's find the value of x. We know that sin(x) = -1/2 and x terminates in the third quadrant. In the third quadrant, both sine and cosine are negative.
Since sin(x) = -1/2, we can conclude that x corresponds to the angle 7π/6 radians or 210 degrees in the unit circle.
Now, let's find the value of tan(2x) using the double-angle formula for tangent:
tan(2x) = (2 * tan(x)) / (1 - tan^2(x))
To find tan(x), we can use the fact that tan(x) = sin(x) / cos(x). Since sin(x) = -1/2 and we know x is in the third quadrant, where cosine is negative, we can write:
cos(x) = -√(1 - sin^2(x))
= -√(1 - (-1/2)^2)
= -√(1 - 1/4)
= -√(3/4)
= -√3/2
So, tan(x) = sin(x) / cos(x)
= (-1/2) / (-√3/2)
= -1 / √3
= -√3 / 3
Now, let's substitute this value into the double-angle formula:
tan(2x) = (2 * (-√3/3)) / (1 - (-√3/3)^2)
= (-2√3) / (1 - 3/9)
= (-2√3) / (1 - 1/3)
= (-2√3) / (2/3)
= (-2√3) * (3/2)
= -3√3
Therefore, the exact value of tan(2x) is -3√3.
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°