If sin(x) = -1/3 and Pi ≤ x ≤ 3Pi/2, then tan(2x) = ?
You need to remember that the tangent function is rational, hence `tan 2x = (sin 2x)/(cos 2x).`
The problem provides the information that the angle x is in quadrant 3, hence `sin xlt0 ; cos xlt 0` .
You need to remember the formula of half of angle such that:
`sin x = sqrt((1-cos 2x)/2)`
Substituting `-1/3` for sin x yields:
`-1/3 = sqrt((1-cos 2x)/2)`
You need to raise to square to remove the square root such that:
`1/9 = (1-cos 2x)/2 =gt 2/9 = 1 - cos 2x`
`cos 2x = 1 - 2/9 =gt cos 2x = 7/9`
You need to use the basic formula of trigonometry to find sin 2x such that:
`sin 2x = sqrt(1 - cos^2 2x)`
`sin 2x = sqrt(1 - 49/81) =gt sin 2x = sqrt(32/81)`
`sin 2x = sqrt32/9`
You need to substitute `sqrt32/9` for `sin 2x` and `7/9` for `cos 2x` in `tan 2x = (sin 2x)/(cos 2x) ` such that:
`tan 2x = (sqrt32/9)/(7/9) =gt tan 2x = (sqrt32/9)*(9/7)`
`tan 2x = sqrt32/7`
Hence, evaluating the tangent of double of angle x yields `tan 2x = sqrt32/7.`
We'll apply the double angle identity for tan (2x):
tan (2x) = 2tan x/[1-(tan x)^2] If x>=pi and x=<3pi/2, then x is located in the 3rd quadrant, where the values of tangent function are positive.
tan x = sin x/cos x
We need to determine cos x from Pythagorean identity:
cos x = -sqrt[1-(sin x)^2]
cos x = -sqrt(1 - 1/9)
cos x = -sqrt (8/9)
Since x is in the 3rd quadrant, the values of cosine function are negative.
tan x = (-1/3)/(-2sqrt2/3)
tan x = 1/2sqrt2
tanx = sqrt2/4
tan (2x) = 2* (sqrt2/4)/(1-1/8)
tan (2x) = [(sqrt2)/2]/(7/8)
tan (2x) = 4sqrt2/7
The requested value of tan(2x) = 4sqrt2/7.