Trig Help???

Question

The point P on the unit circle that corresponds to a real number t is:

find csc(t)
5/7, -2sqrt 6/7

find cot(t)
-sqrt7/4, -3/4

Answer 1

csc is 1/sine, correct? So csc(t) is 1/(-2sqrt6/7), correct? cot is cos/sine, correct? so cot(t) is (-sqrt7/4)/(-3/4), correct? Remember that the coordinates of points on a unit circle are exactly equal to their cosines and sines, respectively.

Answer 2

Unit circle is x² + y² = 1 ie parametrically x = cos(t) y = sint(t)

Now (x, y) ≡ (5/7, -2√6/7)

So cos(t) = 5/7 > 0 and sin(t) = -2√6/7 < 0
Therefore t is in the 4th quadrant (ie for 0 ≤ t ≤ 2π, then specifically 3π/2 ≤ t ≤ 2π)
ie x-coordinate is +ve and y coordinate is -ve

csc(t) = 1/sin(t) = -7/2√6 = -76/12

(x, y) ≡ (-√7/4, -3/4)
So cos(t) = -√7/4 and sin(t) = -3/4
Therefore t is in the third quadrant (ie for 0 ≤ t ≤ 2π, then specifically π ≤ t ≤ 3π/2)
ie x-coordinate is -ve and y coordinate is -ve

Now cot(t) = cos(t)/sin(t) = (-√7/4)/(-3/4)
= √7/3

Answer 3

find csc(t)
5/7, -2sqrt 6/7
csc(5/7) = 1.526536
csc(-2sqrt 6/7) = -1.04078

find cot(t)
-sqrt7/4, -3/4
cot(-sqrt(7/4) = -0.253
cot(-3/4) = -1.0734

Answer 4

There is no need here for a calculator with its decimal approximations; you can be exact.

For point (5/7, -2sqrt(6)/7):

sin(t)= -2sqrt(6)/7
csc(t) = 1/sin(t) = -(7/12)sqrt(6)

For point (-sqrt(7)/4, -3/4):

cot(t) = cos(t)/sin(t) = (-sqrt(7)/4)/(-3/4) = sqrt(7)/3

Answer 5

Points on a circle, with respect to trig. functions are, (cos(x) , sin(x) ).

So, csc(t) for point ( 5/7, -2(6/7)^.5 ).

cos(t)= 5/7
t= cos^-1(5/7)

Then, put that t value into your calculator for csc(t).

The same thing goes for the other point.