let cscx = -5/3 and cosx

Question

let cscx = -5/3 and cosx < 0, find tan x/2

e.mccormick · Answer

So what part of this are you having trouble with?

anonymous · Answer

all of it. haha.
just need help to solve it, please.

e.mccormick · Answer

You have $\cos x < 0$ which means it is negative.  What does that tell you about the possible answers?

anonymous · Answer

negative.

e.mccormick · Answer

Well, it is for tan in the end... so they might or might not.  But what quadrent(s) would the answers be in?

anonymous · Answer

quadrant 3.

e.mccormick · Answer

Is cos x negative anywhere other then Q3?

anonymous · Answer

i have two more that relate to the same question but the two are find secx/2 and find cotx

e.mccormick · Answer

Or are you saying that absed off of both the cscx = -5/3 and cosx < 0 it is Q3.  I was tak9ng it from the start because you said you had trouble with all of it.

e.mccormick · Answer

missed the i there.  lol.  Hit 9.

anonymous · Answer

try working it out, so i can compare it to the answer i got.

e.mccormick · Answer

These are basically about building a triangle on the unit circle.  The sides of the triangle show the relationship of the trig ratios.  Once you know which quad, then it is very easy to sketch and then pull all the answers from the sketch.

e.mccormick · Answer

Well, what answers did you get?

anonymous · Answer

hold on.

e.mccormick · Answer

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e.mccormick · Answer

As you can see from that, the x value will be -4.

e.mccormick · Answer

Things started reloading and lagging, so I never got to see what you had for an answer.  I had this all typed up!  Well, here it is in case you still need that checked.

You have $\cos x < 0$ which means it is negative.  What does that tell you about the possible answers?

sin csc
cos sec

$y=-3$
$r=5$
$x=\sqrt{5^2-(-3)^2}\implies\sqrt{16}\implies\pm4$ and we know it is negative.
$$\begin{matrix}
\sin\theta=\frac{y}{r}&\csc\theta=\frac{r}{y}\
\cos\theta=\frac{x}{r}&\sec\theta=\frac{r}{x}\
\tan\theta=\frac{y}{x}&\cot\theta=\frac{x}{y}
\end{matrix}\ \therefore \
\begin{matrix}
\sin\theta=\frac{-3}{5}&\csc\theta=\frac{5}{-3}\
\cos\theta=\frac{-4}{5}&\sec\theta=\frac{5}{-4}\
\tan\theta=\frac{-3}{-4}&\cot\theta=\frac{-4}{-3}
\end{matrix}$$Now, some of those have two negatives, which would cancel and have an end result of a positive.