Find sen2x, cos2x, and tan2x if =senx−1213 and x terminates in quadrant IV.sen2x = cos2x = tan2x =

The problem seems to be incomplete. You mentioned that sinx = -12/13 and x is in the fourth quadrant, but you didn't specify what you want to find with sin2x, cos2x, and tan2x.

However, I can show you how to find these values given sinx = -12/13 and x is in the fourth quadrant.

Step 1: Find cosx Since x is in the fourth quadrant, cosx is positive. We can find cosx using the Pythagorean identity sin^2x + cos^2x = 1.

cos^2x = 1 - sin^2x cos^2x = 1 - (-12/13)^2 cos^2x = 1 - 144/169 cos^2x = 25/169 cosx = √(25/169) = 5/13

Step 2: Find sin2x, cos2x, and tan2x We can use the double angle formulas:

sin2x = 2sinxcosx sin2x = 2*(-12/13)*(5/13) = -120/169

cos2x = cos^2x - sin^2x cos2x = (5/13)^2 - (-12/13)^2 = 25/169 - 144/169 = -119/169

tan2x = sin2x/cos2x tan2x = (-120/169) / (-119/169) = 120/119

So, sin2x = -120/169, cos2x = -119/169, and tan2x = 120/119.