Question

cot(10pi/3)+(cot(16pi/3))/(cos(17pi/6))

Answer 1

So they gave you a bunch of annoying rotated angles. We need to unwind them so we can see what angle they refer to in one rotation along the unit circle.

\(\large \dfrac{10\pi}{3}\) See how the angle is larger than \(\large 2\pi\)? We want to subtract \(\large 2\pi\) from this angle.

\(\large \dfrac{10\pi}{3}-2\pi \qquad = \qquad \dfrac{10\pi}{3}-\dfrac{6\pi}{3}\qquad = \qquad \dfrac{4\pi}{3}\)

Which tells us,\[\large \cot\left(\frac{10\pi}{3}\right) \qquad = \qquad \cot\left(\frac{4\pi}{3}\right)\]

Answer 2

btw theyre asking for the answer in improper form

Answer 3

Hmm I'm not sure what that means. Do they just mean, don't round your answer to a decimal?

Answer 4

im guessing just get the answer without further simplifying

Answer 5

Nah I don't think that's what they mean :)
You know that 4pi/3 is a special angle. They won't expect you to remember 10pi/3.
You have to do a little work to refer it back to your special angle.

Answer 6

\[\large \cot\left(\frac{4\pi}{3}\right) \qquad = \qquad \frac{1}{\sqrt3}\]
Right?

I mean, if you want to try and do this without simplifying you can, but that seems rather silly. :D

Answer 7

ok confused at your answer of it equalling 1/sqrt3

Answer 8

If you don't remember cotangent, that's ok. We can rewrite it in terms of sines and cosines.

\[\large \cot x=\frac{\cos x}{\sin x} \qquad \rightarrow \qquad \frac{\cos\left(\dfrac{4\pi}{3}\right)}{\sin\left(\dfrac{4\pi}{3}\right)}\]Do you remember these special angles?

Answer 9

yes

Answer 10

hello?

Answer 11

Yes you do recall what those special angles give you?
So are you able to simplify the fraction? :o