# pre-calculus

tanx= -12/5 x in quadrant 2
find;
sin2x=
cos2x=
tan2x=

1. Damon

draw a sketch of x y axis system

x = -5
y = 12
hypotenuse = 13

sin x = 12/13
cos x = -5/13
tan x = -12/5
now use 2x formulas
for example
sin 2x = 2 sin x cos x = 24/13 * -5/13
= -120/169

## Similar Questions

1. ### pre-cal

find all solutions in interval [0,2pi] tan²x = -3/2 secx This question may look tricky, but its actually quite simple. First: rearrange the pythagorean identity tan²x+1=sin²x so that it reads tan²x=sin²x-1 Then replace the tan²x …
2. ### trig

sinx = 4/5 and x terminates in Quadrant II Find sin2x and cos2x How to get the answers, which are sin2x = -24/25, cos2x = -7/25?
3. ### Trigonometry

I can't remember what formula to use for these. Note- the X's may be replaced by a theta symbol. I just didn't have one on my keyboard. Use the given information to find sin2X, cos2X and tan2X if 0< X <pi/2 1. sinX=12/13 2. cosX=3/5 …
4. ### trig

Verify the trigonometric identity: [(1–sin²x)/sin²x]–[(csc²x–1)/cos²x]= -tan²x I still can't figure this out.
5. ### Calculus

Solve identity, (1-sin2x)/cos2x = cos2x/(1+sin2x) I tried starting from the right side, RS: =(cos²x-sin²x)/(1+2sinxcosx) =(cos²x-(1-cos²x))/(1+2sinxcosx) and the right side just goes in circle. May I get a hint to start off?
6. ### Pre Calculus

Establish the following identity Sin2x/1+cos2x=tanx
7. ### pre-calc

tanx= -12/5 x in Q 2. find sin2x= cos2x= tan2x=