$\displaystyle cos 190^o$

$\displaystyle cot 290^o$

Printable View

- Jul 28th 2009, 05:43 PMcrosser43Write using a positive acute angle?
$\displaystyle cos 190^o$

$\displaystyle cot 290^o$ - Jul 28th 2009, 05:59 PMpickslides
$\displaystyle \cos(190) = -\cos(10)$

$\displaystyle \cot(290) = \frac{-1}{\tan(50)}$ - Jul 28th 2009, 06:04 PMcrosser43
thanks, but how do you come up to that?

- Jul 28th 2009, 06:19 PMANDS!
We first need to know what an "acute angle" is. According to the internets:

$\displaystyle \mbox{An acute angle is one which is less than} \ 90^\circ$

Thus, we need to figure out a way to convert these angles into something that is less than $\displaystyle 90^\circ$. Hopefully you are familiar with reference angles (which are acute angles themselves), as that makes this much easier.

With $\displaystyle \cos(190^\circ)$, we are dealing with an angle in the third quadrant. If we take $\displaystyle 190^\circ$ we find that we go 10 degrees past the X-axis, landing in the third quadrant. Cosine in the third quadrant is negative. However, the only angles (general angles) that make cosine NEGATIVE are NOT acute angles. If we just wrote $\displaystyle \cos(10^\circ)$ we would get a POSITIVE number, when $\displaystyle \cos(190^\circ)$ is NEGATIVE (check it on your calculator). Thusly, we need to figure out some way of turning that number negative.

Now, since cosine is an EVEN function, slapping a negative on to the angle measure wont work (look at a unit circle; if you go 10 degrees clockwise and end up in the 4th quadrant, any cosine values you take will STILL be positive). The only other option is to take the NEGATIVE of the entire function (not just the variable). And thus we end up with $\displaystyle -\cos(10^\circ)$.

Does this make sense? - Jul 28th 2009, 06:20 PMpickslides
$\displaystyle \cos(180 +\theta)=-\cos(\theta)$

$\displaystyle \tan(360 +\theta)=-\tan(\theta)$

where $\displaystyle \tan(\theta) = \frac{1}{\cot(\theta)}$