SOLUTION: Use the given information to find cos 2x, sin 2x, and tan 2x. cosx=sqrt3/7 and 3&#960;/2 < x < 2&#960; since I have cos, I created a triangle using sqrt3 as the opposite side.

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Question 737089: Use the given information to find cos 2x, sin 2x, and tan 2x.
cosx=sqrt3/7 and 3π/2 < x < 2π
since I have cos, I created a triangle using sqrt3 as the opposite side. and 7 as the adjacent side. Then I used the Pythagorean theorem to find the hypotenuse.
hypotenuse^2=(sqrt3)^2+(7)^2
hypotenuse^2=3+49
hypotenuse^2=52
hypotenuse=sqrt52
Then to find sin2x, I used sin2x=2sin(xcos(x), so...
sin2x=2sin(xcos(x)
= 2(sqrt3/sqrt52)(sqrt3/7)
= 2 (sqrt9/7sqrt52)
= 3/7sqrt13
but that was incorrect and I can't go any further in the problem with an incorrect sin2x. Please assist me with where I went wrong or how to go about finding this answer. thanks!

Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website!
cos x =
square both sides
cos^2 x=3/49
1-cos^2 x=1-3/49
sin^2 x = 46/49

cos 2x = 2 cos^2 x -1
=2*3/49-1
=6/49-1
=-43/49
Cos 2x= -43/49
Sin 2x = 2sin x . cos x
=
=
Tan (2x) = sin (2x)/cos (2x)

SOLUTION: Use the given information to find cos 2x, sin 2x, and tan 2x. cosx=sqrt3/7 and 3&#960;/2 < x < 2&#960; since I have cos, I created a triangle using sqrt3 as the opposite side.

SOLUTION: Use the given information to find cos 2x, sin 2x, and tan 2x. cosx=sqrt3/7 and 3π/2 < x < 2π since I have cos, I created a triangle using sqrt3 as the opposite side.