Pre-Cal Homework Plzz Help?

Question

Hi, i have pre cal homework today. It seems that i cant figure out a problem. It tells me to find the values of the six trignometric functions of the theta. So what i have on the sheet of problems is

Function Value
cotangent theta is undefined

Constraint
pi/2 less than or equal to theta less than or equal to 3pi/2

If you know that answer can you please tell me what is (x,y) and what quadrant is the triangle in.

Answer 1

Ok, let's see...

cot(theta) = 1/tan(theta)

tan(theta) range:

Value: 0 for theta = 0 deg
Value: inifinity for theta = 90 degrees

So, for cot(theta) the values are reversed

Value: infinity for theta = 0 deg
Value: 0 for theta = 90 deg (or pi/2)

The same analysis is valid for theta = 3pi/2 (or -90 deg in the fourth quadrant).

Answer 2

If cot(theta) is undefined, that probably means the theta = 0, pi,..,n*pi.

Within your constraints, theta must be pi, since 0 < pi/2, and 2*pi > 3*pi/2. The remaining functions are:

sin(pi) = 0, cos(pi) = -1, tan(pi) = 0 cosec(pi) = undef, sec(pi) = -1

I don't know what you mean by x,y; and an angle of pi (180 deg) does not form a triange, but lies on the negative x-axis between quadrants IIi and IV

Answer 3

@ theta = pi/2 => (x,y) = (0,1), cotangent(theta) = x/y = 0

theta between pi/2 and pi -> QUADRANT II (2)

@ theta = pi => (x,y) = (1,0), cotangent(theta) = x/y = undefined (division by zero)

theta between pi and 3pi/2 -> QUADRANT III (3)

@ theta = 3pi/2 => (x,y) = (0,-1), cotangent(theta) = x/y = 0

Defn:
sqrt -> SQUARE ROOT FUNCTION
^ -> POWER (x^2 = [x squared])
sine(theta) = y/[sqrt(x^2 + y^2)]
cosine(theta) = x/[sqrt(x^2 + y^2)]
tangent(theta) = sine/cosine = y/x

Answer 4

not taking pre calc yet