Question

use the given values to evaluate (if possible)all six trig functions.
cot theta=-5
sin theta =(square root 26)/26

Answer 1

@johnweldon1993

Answer 2

Well you know that

\[\large \sin(x) = \frac{\text{Opposite}}{\text{Hypotenuse}}\]

so

Answer 3

\[\large \cot(x) = \frac{1}{\tan(x)} = \frac{1}{\frac{sin(x)}{cos(x)}} = \frac{cos(x)}{sin(x)}\]

Answer 4

We also know that sin = √26/6 and that cot = -5 so

\[\huge \frac{cos(x)}{\frac{\sqrt{26}}{26}} = -5\]
\[\huge \frac{26cos(x)}{\sqrt{26}} = -5\]
\[\large 26cos(x) = -5 \times \sqrt{26}\]
\[\large cos(x) = \frac{-5\sqrt{26}}{26}\]

Now from that we know \(\large cos(x) = \frac{\text{adjacent}}{\text{hypotenuse}}\) so that means our hypotenuse would be 26 (and you see above that it is...and our missing side is \(\large -5\sqrt{26}\)

Answer 5

And now you have your 3 sides to use for the trig functions...which I'm assuming you know