Precal trigonometry, moving along the unit circle

Question

Starting at the reference point (1,0) on the unit circle, a person traces a line counterclockwise along that circle a distance of x landing at the terminal point (-20/29, 21/29). Calculate all of the trigonometric functions evaluated at x, and all the trigonometric functions evaluated at 2x.

Answer 1

sin(x) = opp/hyp = (21/29)/1 = 21/29

cos(x) = adj/hyp = (-20/29)/1 = -20/29

tan(x) = opp/adj = (21/29)/(-20/29) = -21/20

csc(x) = 1/sin(x) = 29/21

sec(x) = 1/cos(x) = -29/20

cot(x) = 1/tan(x) = -20/21

For 2x:

Use trig identities:

sin(2x) = 2sin(x)cos(x) = 2(21/29)(-20/29) = -840/841

cos(2x) = 2cos²(x) – 1 = 2(-20/29)² - 1 = 800/841 - 1 = -41/841

tan(2x) = 2tan(x)/(1 - tan²(x)) = 2(-21/20)/(1 - (-21/20)²) = (-21/10)/(-41/400) = 840/41

csc(x) = 1/sin(x) = -841/840

sec(x) = 1/cos(x) = -841/41

cot(x) = 1/tan(x) = 41/840

Answer 2

This would put x in the second quadrant of the unit circle. x in the problem is really the angle. Let's call it θ instead so not to confuse it with the x-y coordinates. The distance is actually the arc rθ, but since it is a unit circle(r=1), distance = θ. (x,y) = (-20/29, 21/29)

sin(θ) = y/r = 21/29

cos(θ) = x/r = -20/29

tan(θ) = y/x = (21/29)/(-20/29) = -21/20

sec(θ) = 1/cos(θ) = -29/20

csc(θ) = 1/sin(θ) = 29/21

cot(θ) = 1/tan(θ) = -20/21

For the second part, use the double angle formulas and I'll let you finish the math.

sin(2θ) = 2sin(θ)cos(θ) = 2(21/29)(-20/29)

cos(2θ) = cos²(θ) - sin²(θ) = (-20/29)²-(21/29)²

tan(2θ) = sin(2θ)/cos(2θ)

sec(2θ) = 1/cos(2θ)

csc(2θ) = 1/sin(2θ)

cot(2θ) = cos(2θ)/sin(2θ) = 1/tan(θ)