Question

Find the exact values of the six trigonometric functions of θ if θ is in standard position and the terminal side of θ is in the given quadrant and satisfies the given condition.
III; bisects the quadrant

Answer 1

do you mean bisects the quadrant 3?

Answer 2

if so then the values of the six will be
\[\sin \theta = -\sqrt{2}/2\]
\[\cos \theta=-\sqrt{2}/2\]
\[\tan \theta=1\]
\[\cot \theta=1\]
\[\csc \theta=-\sqrt2\]
\[\sec \theta=-\sqrt2\]

Answer 3

yep, thanks. explain pls how did you find it?

Answer 4

ok first let me sketch this

Answer 5

since the angle formed in quadrant III is 90 degrees, then if we bisect that angle we'll get half of that which is 45 degrees....
so we know that in a right triangle with 45 degrees, the sides will have a ratio of 1:1:square root of 2
so let's suppose our hypotenuse is sqrt of 2 and the legs have measure of 1. Since we're in quadrant 3 both legs will have measure of -, (because at quadrant 3, x and y are both negative) 1 but the hypotenuse will still be sqrt2... so we'll have

using this info, you can now compute for the 6 trig. functions of theta..
so sin theta=-1/sqrt2=-sqrt(2)/2
cos theta=-1/sqrt2=-sqrt(2)/2
tan theta=sin theta/cos theta <a trigonometric identity
since sin theta=cos theta in this case,
tan theta=1
cot theta=1/tan theta=1/1=1
sec theta= 1/cos theta=1/(-1/sqrt2)=-sqrt2
csc theta=1/sin theta=1/(-1/sqrt2)=-sqrt2