Sec x = -13/5, pi/2 < x < 2 pi find cos2x, sin2x and tan2x

Question

I am absolutely terrible with trig functions and I am not sure how to even start

Answer 1

look at this as a right triangle. A common right triangle is 5, 12 , 13 where 13 is hypotenuse, 5 is adjacent side, 12 is opposite.

 

secx = 1/cosx = -13/5. Thus cosx = -5/13. = adjacent/hypotenuse 

if you look at quadrants cosx is only negative on the second and third quadrant (I.e pi/2 < X < 3Pi//2)

 

what is sinx? Well sinx = opp/hyp = 12/13 if I'm second quadrant and -12/13 if in 3rd.

 

use angle identity sin2x = 2sinxcosx. Plug in values of sinx and cos X. You will have either plus or minus 120/169 I believe.

 

use angle identity cos2x = 1-2(sinx)^2. You have sinx So plug in. You should have like 169/169-2*144/169 = -119/169 I believe.

 

use tan2x = sin2x/cos2x = plus or minus 120/119 I believe.