Remaining Trig Functions of Theta?

Question

Find the exact value of each of the remaining trig functions of theta.

Let t = theta

(1) cos(t) = 3/5, where t lies in quadrant 4

(2) sec(t) = -2, where tan(t) > 0

Answer 1

cos = adj / hyp
sin = opp / hyp
tan = opp / adj
sec = hyp / adj
csc = hyp / opp
cot = adj / opp

1)
cos t = 3 / 5 = adj / hyp
4th Quad means sin will be negative.

Using pythagorean theorem, you can find the missing side (opp).
(Hyp)^2 = (adj)^2 + (opp)^2
you get opp = 4, but in this case, because it's quad 4,
opp = -4.

So, you have
adj = 3
opp = -4
hyp = 5

Just fill those in to find the other trig functions. (I listed them above.)

2)
sec t = -2 / 1 = hyp / adj
tan > 0 is in the 1st and 3rd quad. Since sec is negative, then it has to be in the third quad. So, both opp and adj are negative values. (Hyp is always positive.)

Use pythagorean theorem to find opp
(hyp)^2 = (adj)^2 + (opp)^2

Opp = -sqrt 3
Adj = -1
Hyp = 2

Use those values to find the other trig functions.

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

1)

In this from cos t = 3/5 = base/hypotenuse

draw a rt angle triangle with b=3 hyp=5
using pythagoras theorem, third side perpendicular(p) = 4.

so, sin t = -4/5 ( - because 4th quad)
tan t = -4/3
cot t= -3/4
cosec t= -5/4
sec t= 5/3

2) se

Answer 4

Question 1
cos θ = 3 / 5 is + ve and therefore θ lies in 1st and 4th quadrants.
θ = 53.1° and 360 - 53.1°
θ = 53.1° , 307 °

Question 2
1 / cos θ = - 2
- 1/2 = cos θ
θ is - ve and therefore θ lis in 2nd and 3rd quadrants.
θ = 180 - 60° , θ = 180 + 60°
θ = 120° , x = 240°
But condition of question is that tan θ is + ve.
Therefore accept answer of θ = 240° only.

Answer 5

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