Trig Review

**DUKENUKEM** · August 4th 2008, 08:17 PM

Hello, i need help in refreshing of memory with these 4 different types of equations. It's been about a year since i've taken trig and school just started calculus class today and i am lost.

Here is 1 example of each section that i need help with:

Find exact value of:
$sin(35(Pi)/6)$

Evaluate each for 0 < 0 < 2(pi):
$tan^-1(-1) $
Find the domain and range:
$y = ((2x-3)/(x+1))$

The other one is a word problem, where as again, i have no clue where to start.
You are given the following vertices of [TRIANGLE]ABC, A(-1,-1), B(3,5), and C (5, -3). State the process to find point of intersection of the median from C to AB and the median form B to AC.

I am pretty stressed right now considering these 20 problems are gonna kill me

, any help would be greatly appreciated

**Chris L T521** · August 4th 2008, 09:29 PM

Originally Posted by DUKENUKEM

Hello, i need help in refreshing of memory with these 4 different types of equations. It's been about a year since i've taken trig and school just started calculus class today and i am lost.

Here is 1 example of each section that i need help with:

Find exact value of:
$sin(35(Pi)/6)$

These can be tricky at first, but you'll get this quickly.

Take note that we are dealing with angles that are integer multiples of $2\pi$ .

With little effort, we can see that $\tfrac{35}{6}\pi=\tfrac{36}{6}\pi-\tfrac{1}{6}\pi=6\pi-\tfrac{1}{6}\pi$

Since $6\pi$ is an integer multiple of $2\pi$ , we can (in essence), ignore it. Thus, what we are asked to find is $\sin\left(-\tfrac{1}{6}\pi\right)=-\sin\left(\tfrac{1}{6}\pi\right)=\color{red}\boxed {-\tfrac{1}{2}}$ .

I hope this makes sense!

Evaluate each for 0 < 0 < 2(pi):
$tan^-1(-1) $

When we evaluate $\tan^{-1}(-1)$ , we come up with many values. However, we're given a restriction on $\vartheta$ :

$0\leq\vartheta\leq 2\pi$

So, we see that $\tan^{-1}(-1)=\tfrac{3}{4}\pi, \ \tfrac{7}{4}\pi, \ \tfrac{11}{4}\pi, \ \tfrac{15}{4}\pi, \ ...$

Taking into consideration the restriction of $\vartheta$ , we see that the only possible solutions woud be $\color{red}\boxed{\tfrac{3}{4}\pi, \ \tfrac{7}{4}\pi}$

I hope this makes sense!

Find the domain and range:
$y = ((2x-3)/(x+1))$

At $x=-1$ , we have a horizontal asymptote. Thus, the domain of y would be $\color{red}\boxed{\left(-\infty,-1\right)\text{ and }\left(-1,\infty\right)}$

If we evaluate $\frac{2x-3}{x+1}$ at large positive and negative values, we see that the function starts to get closer and closer to a value of $y=2$ . Since there was a discontinuity at $x=-1$ , we can determine that at $y=2$ , there is a discontinuity as well. Thus, we can claim the range to be $\color{red}\boxed{\left(-\infty,2\right)\text{ and }\left(2,\infty\right)}$

I hope this makes sense!

The other one is a word problem, where as again, i have no clue where to start.
You are given the following vertices of [TRIANGLE]ABC, A(-1,-1), B(3,5), and C (5, -3). State the process to find point of intersection of the median from C to AB and the median form B to AC.

I'm trying to figure out what you mean by this. Could you state the problem as it is written? I'm sure I'd be able to do this with a little clarification.

I hope all that I have done so far makes sense. If you have questions, please ask.

--Chris

**DUKENUKEM** · August 5th 2008, 07:09 PM

i understand the first and 3rd now

and i figured out the word problem by finding the explaining to find the midpoints that cross the requested sides.

but the second one is still roaming around in my head...

how did you get from
$ \tan^{-1}(-1) $
to

$ \tan^{-1}(-1)=\tfrac{3}{4}\pi, \ \tfrac{7}{4}\pi, \ \tfrac{11}{4}\pi, \ \tfrac{15}{4}\pi, \ ... $

i've remember vaguely doing something similar to this, however, it's just not coming back to me.

Thanks for the help on the others

**mr fantastic** · August 5th 2008, 07:58 PM

Originally Posted by DUKENUKEM

i understand the first and 3rd now

and i figured out the word problem by finding the explaining to find the midpoints that cross the requested sides.

but the second one is still roaming around in my head...

how did you get from
$ \tan^{-1}(-1) $
to

$ \tan^{-1}(-1)=\tfrac{3}{4}\pi, \ \tfrac{7}{4}\pi, \ \tfrac{11}{4}\pi, \ \tfrac{15}{4}\pi, \ ... $

i've remember vaguely doing something similar to this, however, it's just not coming back to me.

Thanks for the help on the others

$\tan \theta = -1 \Rightarrow \theta = \tan^{-1} (-1)$ .

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