Remaining Trig Functions of Theta?

Question

Remaining Trig Functions of Theta.

(1) sin(theta) = -(2/3), where pi < theta < (3pi/2).

(2) cos(theta) = 4/5, where 270 degrees < theta < 360 degrees.

Answer 1

1) sin t = -2/3
cos t = - sqr(5)/3
tan t = 2/sqr(5)
cot t = sqr(5) /2
sec t = - 3 / sqr(5)
csc t = -3 / 2

2) cos t = 4/5
sin t = - 3/5
tan t = - 3/4
cot t = - 4/3
sec t = 5/4
csc t = - 5/3

Answer 2

Sorry about the delay, I see you already have reasonable answers. Apparently one of the reference links was incomplete and security wouldn't let the submission through! You can find some good information at www.clarku.edu under trig functions *****original answer***** Use the unit circle (r=1) to visualize the functions with sin [t] = y/r cos [t] = x/r tan [t] = y/x and their geometric inverses csc [t] = r/y sec [t] = r/x cot [t] = x/y then using cos[t] = 3/5 = 0.6, the the unit circle, and the Pythagorean theorem we get y = sqrt [1^2 - 0.6^2] = sqrt [1-0.36] = sqrt [0.64] = +/- 0.8 but quadrant 4 is specified so y = -0.8 therefore, in decimal terms: sin [t] = y/r = -0.8/1 = -8/10 = -0.8 cos [t] = x/r = 0.6 (given) tan [t] = y/x = -0.8/0.6 = -4/3 = -1.333... csc [t] = r/y = 1/-0.8 = -1.25 sec [t] = r/x = 1/0.6 = 10/6 = 1.66... cot [t] = 1/sin [t] = x/y = -3/4 = -0.75 You should be able to do the next problem similarly, using values of t in quadrant 3 (where sec t is neg and tan t is > 0)

Answer 3

Question 1
sin Ө = - 2 / 3 ,
Ө in 3rd quadrant
Ө = 221 . 8°
Ө = 3.86 radians
cos Ө = - 0.745
tan Ө = 0.894

Question 2
Ө in 4th quadrant
Ө = 323.1°
sin Ө = - 0.6 = - 3 / 5
tan Ө = - 0.75 = - 3 / 4

Answer 4

sin(theta)= - 2/3 cosec(theta)= -3/2
cos(theta) = 4/5 sec(theta)= 5/4
tan(theta)= sin(theta)/cos(theta) = -5/6
cot(theta) = 1/tan(theta) = - 6/5