Question

Determine the Value of all six trigonometric functions of theta when given?
I need help. This is for my midterm review, if anyone is willing to help, I'd appreciate it.

a) sin theta= 3/5 and theta is in quadrant 1.
b) tan theta= -2 and theta is in quadrant II
c) sec theta = 2 radical 3/3 and theta is in quadrant II
d) csc theta= -2/3 and theta is in quadrant III.

I need to find all six trigonometric functions for each one ( sin, cos, tan, csc, sec, cot)

Answer 1

With each function given, you have two sides of a right triangle. Using the pyth theorm, you can find the other side, and thus, the other functions. We will be happy to check your work.

For example, c) If secTheta = 2 sqrt (1/3), then the sides of the triangle are

opposite figure out
adjacent sqrt 3
hypo= 2

and the opposite will be

4=3+opposite^2 or opposite is +-1, Now in quadrant two, that means sec is negative, so it should have been -2sqrt(1/3), and the opposite (vertical) is 1. From that you get the six functions.

Answer 2

I will do c. You try the rest.
Oh no I won't --> sign of sec is negative in quad 2. I think typo
I will do d.
d) csc T = 1/sin T = -2/3
Hey! what is going on here?
the absolute value of sin T may not be greater than 1. It can not be -3/2
Either we are looking at typos or someone is playing games with you.
I will do b I guess
b) draw triangle in quad 2
-1 along -x axis
+2 up at x = -2
now hypotenuse = sqrt (-1^2 + 2^2) = sqrt 5
so
sin = 2/sqrt 5
cos = -1/sqrt 5 = -(1/5)sqrt 5
tan = -2/1
csc = 1/sin = (1/2)sqrt 5
sec = 1/cos = -sqrt 5
cot = 1/tan = -1/2

Answer 3

prove the identities (theta + phi), cos theta = -1/3 in Quadrant III, sin theta = 1/4 in quadrant II