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mattlai7
if cos(t)=-1/5 and pi<t<3pi/2. find csc(t)
|dw:1362772905106:dw|Understand why I drew that triangle in the third quadrant?
|dw:1362773055192:dw| They told us that the angle t in within this interval.\(\large \pi<t<3\pi/2\). So we draw a line in the third quadrant and form a triangle from it.
ok now i understand that part
|dw:1362773145892:dw|Let's look at just the triangle now. Don't worry about the fact that I labeled the angle \(\large t'\). That's not super important. Just think of it as \(\large t\).
\(\large \cos t=-\dfrac{1}{5}=\dfrac{adjacent}{hypotenuse}\) So how would we label these sides? HINT: We always let the hypotenuse be positive.
Ok looks good! Just need to find the missing side using the `Pythagorean Theorem` and then we just have one small step after that.
opposite will equal 4?
sorry did an error......missing side equals to \[+/- 2\sqrt{6}\]
Ok looks good. Now remember that we drew our triangle in the `third` quadrant. So for that OPPOSITE leg, did we move UP or DOWN? Do we want the positive or negative answer?
we want a negative answer because tan is only positive in the 3rd Q
Haha that's an interesting way to think about it XD I was thinking about the fact that we moved downward in the y direction to make the opposite length. So it would be negative. But whatever works for you! :D|dw:1362773803586:dw|
Ok good. Everything is labeled correctly. Just have to find \(\large \csc t\) now! :)
hypotenuse over opposite
For your final answer it might be a good idea to `rationalize` your answer. Get the irrational number out of the denominator. Multiply the top and bottom by sqrt6.
a little confused on the denominator part....not sure if its 12 or not
\[\large 2 \cdot \sqrt6 \sqrt6 \qquad = \qquad 2\cdot6 \qquad = \qquad 12\]Hmm yah everything looks ok there!
thanks for the help some time a 8am class only aborbs so much in an hour and with 300 classmates lol