1. ## Trigonometric Identities

Hi, I was wondering if anyone could help me with this problem? It's the last once I need to complete in order to get an A on the assignment. I'm pretty sure it has to do with trigonometric identities, I just don't know how to apply it out properly. I'd really appreciate the help, thank you in advance.

2. Hello, DesiKid89!

Suppose $\displaystyle \sin t = \frac{2}{11}$ and $\displaystyle \tan t < 0$

Find each of the following . . .

Sine is positive in Quadrants 1 and 2.
Tangent is negative in Quadrants 2 and 4.
. . Hence, $\displaystyle t$ is in Quadrant 2.

$\displaystyle a)\;\;\sin(-t)$
$\displaystyle \sin(-t) \:=\:-\sin(t) \:=\:\boxed{-\frac{2}{11}}$

$\displaystyle b)\;\;\cos(6\pi-t)$
We have the identity: .$\displaystyle \sin^2t + \cos^2t \:=\:1\quad\Rightarrow\quad\cos t \:=\:\pm\sqrt{1-\sin^2t}$

We are given: .$\displaystyle \sin t \:=\:\frac{2}{11}$

Hence: .$\displaystyle \cos t \;=\;\pm\sqrt{1 - \left(\frac{2}{11}\right)^2} \;=\;\pm\sqrt{\frac{117}{121}} \;=\;\pm\frac{3\sqrt{13}}{11}$

Since $\displaystyle t$ is in Quadrant 2: .$\displaystyle \cos t \;=\;-\frac{3\sqrt{13}}{11}$

Back to the problem:

$\displaystyle \cos(6\pi-t) \;=\;\cos(-t) \;=\;\cos(t) \;=\;\boxed{-\frac{3\sqrt{13}}{11}}$

$\displaystyle c)\;\;\tan t$
$\displaystyle \tan t \;=\;\frac{\sin t}{\cos t} \;=\;\frac{\frac{2}{11}}{-\frac{3\sqrt{13}}{11}} \;=\;-\frac{2}{3\sqrt{13}} \;=\;\boxed{-\frac{2\sqrt{13}}{39}}$

$\displaystyle d)\;\;\csc(t + 6\pi)$
$\displaystyle \csc(t + 6\pi) \;=\;\csc t \;=\;\frac{1}{\sin t} \;=\;\frac{1}{\frac{2}{11}} \;=\;\frac{11}{2}$

Are you getting the idea?
Can you do the rest?