Find sin⁡(2x), cos⁡(2x), and tan⁡(2x) for the given situation: cosx= -1/3,π/2 < x < π

Question

stuck on how to figure this out

Answer 1

cosx = -1/3

use a calculator with an inverse trig function

x = about 109.5 degrees which is in quadrant II

cos2x = cos219 =about -0.777

sin2x = sin219 =about -0.629

tan2x = tan219 =about 0.810

or another method with exact solutions

cosx = -1/3

sinx = sqr(1-(1/3)^2) = sqr(8/9) = (2/3)sqr2

sin2x = 2sinxcosx = 2(2/3)sqr2(-1/3) = exactly -(4/9)sqr2 = about -0.629

cos2x = cos^2x - sin^2x = (1/9) - (4/9)2 = 1/9-8/9 = exactly -7/9 = about -0.777

tan2x = sin2x/cos2x = -(4/9)sqr2/(-7/9) = -(4/9)sqr2)(-9/7) = exactly (4/7)sqr2 = about 0.81

above uses some trig identities that you may need to look up

sin2x = 2sinxcosx

cos2x = cos^2(x) - sin^2(x)

and

sinx = + or - square root of (1-cos^2(x)), using the positive square root as x is in quadrant 2

pi/2 < x < pi is the same as 90 degrees < x < 180 degrees, which means quadrant 2