SOLUTION: If tan x = -1/3 and cos x >0, then.... sin 2x = ? cos 2x = ? tan 2x = ? and If cos x = 4/5 and csc x<0, then.. sin 2x = ? cos 2x = ? tan 2x = ? For every tan

Question 671746: If tan x = -1/3 and cos x >0, then....
sin 2x = ?
cos 2x = ?
tan 2x = ?

and

If cos x = 4/5 and csc x<0, then..
sin 2x = ?
cos 2x = ?
tan 2x = ?
For every tan 2x do I just use the formula sin theta / (1 + cos theta)?
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
If tan x = -1/3 and cos x >0, then
By quadrants:
cos + - - +
tan + - + -
--> Q4
tan(x) = -1/3 --> terminal point (3,-1)
r = sqrt(10)
---------
sin(x) = -1/sqrt(10) = -sqrt(10)/10
cos(x) = 3/sqrt(10) = 3sqrt(10)/10
-------------
sin 2x = 2sin(x)cos(x) = -3/5
cos 2x = sqrt(1 - sin^2) = 4/5
tan 2x = sin(2x)/cos(2x) = -3/4
and
If cos x = 4/5 and csc x<0, then
-----------
By quadrants:
cos + - - +
csc + + - -
--> Q4
r = 5, x = 4, y = -3
Terminal point (4,-3)
----------
sin(x) = y/r = -3/5
sin 2x = 2sin(x)cos(x) = -24/25
cos 2x = sqrt(1 - sin^2(2x)) = 7/25
tan 2x = sin(2x)/cos(2x) = -24/7
For every tan 2x do I just use the formula sin theta / (1 + cos theta)?
Not sure what you mean by that.
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