Problems which I cannot for the life of me figure out why I got wrong/partially incorrect:
A: Find all six trigonometric functions of θ if the given point is on the terminal side of θ. (Assume that a is a positive number. If an answer is undefined, enter UNDEFINED.)
(−5a, −12a)
Result: http://prntscr.com/8diac1
B: Use your calculator to find sin θ and cos θ if the point (9.37, 7.05) is on the terminal side of θ. (Round your answers to four decimal places.)
Result: http://prntscr.com/8dia9h
C: Indicate the quadrants in which the terminal side of θ must lie under the following conditions. (Select all that apply.)
cos θ and cot θ have the same sign
Result: http://prntscr.com/8diaew

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- kittiwitti1

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- anonymous

|dw:1441605814325:dw|
So we get
\[\sin(\theta)=\sin(180+x)=-\sin(x)\]
\[\cos(\theta)=\cos(180+x)=-\cos(x)\]\[\tan(\theta)=\tan(180+x)=\tan(x)\]
from the triangle we can get
\[\sin(x)=\frac{-12a}{\sqrt{(-12a)^2+(-5a)^2}}=\frac{-12a}{\sqrt{144a^2+25a^2}}=\frac{-12a}{13a}=-\frac{12}{13}\]
Similarly find cos(x) and tan(x) and use them to find sin(theta), cos(theta) and tan(theta)
reciprocal for cosec(theta) sec(theta) and cot(theta)

- kittiwitti1

Which problem are you doing? (Sorry I'm multitasking) e_e

- anonymous

Do you have a figure?Talking about problem 1

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## More answers

- kittiwitti1

I'm only wondering where I went wrong, I don't need the whole process explained e_e;;;

- kittiwitti1

Besides they said the sin was correct...
My issue is how the tan/cot are incorrect.

- anonymous

I see your problem now, what you did was, you found the ratios
\[\sin(x),\cos(x),\tan(x)\]
BUT you're suppose to find
\[\sin(\theta),\cos(\theta),\tan(\theta)\]
and NOT
\[\sin(x),\cos(x),\tan(x)\]
what I mean by x, refer to the figure

- kittiwitti1

I don't see what you mean by x... yes I looked at the figure. x_x;

- anonymous

You did something like |dw:1441606633005:dw|

- kittiwitti1

Yes but aren't the trig values the same?

- anonymous

Nope!! It makes a slight difference

- kittiwitti1

It does?
*trig function values

- anonymous

Refer to these formulae I've given below
\[\sin(\theta)=\sin(180+x)=-\sin(x)\]\[\cos(\theta)=\cos(180+x)=-\cos(x)\]\[\tan(\theta)=\tan(180+x)=\tan(x)\]

- anonymous

As you can see sign changes for sine and cosine but remains same for tangent

- anonymous

Let's find your tan theta, since it's seem to be giving you problem
\[\tan(\theta)=\tan(x)=\frac{-12a}{-5a}=\frac{12}{5}\]

- kittiwitti1

Er. Yes...

- anonymous

Actually you've mixed your adjacent and opposite for tangent

- anonymous

that's why you're getting wrong answer

- anonymous

and if your tangent is wrong then obviously cotangent will also be wrong

- anonymous

notice you answers you've done
\[\frac{5}{12}\]
when it should be
\[\frac{12}{5}\]

- anonymous

@kittiwitti1 you there ??

- kittiwitti1

Sorry, working on other problems... What'd I miss?

- anonymous

for problem 1 actually you've exchanged your adjacent and opposite accidentally

- kittiwitti1

-facedesks- Dx
I feel dumb.
Okay, thanks; what about the other two?

- anonymous

problem 3)
Quadrant 2, both cos and cot have negative sign so that's ok
Quadrant 3, cos is negative while cot is POSITIVE, so that's wrong
how about Quadrant 1?both cos and cot, infact all functions are positive!

- kittiwitti1

B: http://prntscr.com/8dia9h
C: http://prntscr.com/8diaew

- anonymous

|dw:1441607496705:dw|

- freckles

\[\sin(\theta)=\frac{y}{r} \text{ and } \cos(\theta)=\frac{x}{r} \\ \text{ and you have } r=\sqrt{x^2+y^2} \\ \text{ you can use this \to help you evaluate } \sin(\theta) \text{ and } \cos(\theta)\]

- kittiwitti1

Oh... I seem to have attracted quite the audience... sorry for previous "mass-tags" ^^;

- anonymous

|dw:1441607639710:dw|

- kittiwitti1

Wait what are you going on about lol

- anonymous

After->1st quadrant
School->2nd quadrant
To->3rd quadrant
College->4th quadrant

- anonymous

It's an easy way to remember

- anonymous

like soh cah toa

- kittiwitti1

Yes but what does that have to do with problem B o_o (I mean thanks for the info but I don't see the connection)

- anonymous

In problem 3 quadrant is wrong look at it again

- anonymous

3rd quadrant

- anonymous

is wrong

- anonymous

I told you about that just in case you have any trouble remembering what is positive in which quadrant

- kittiwitti1

Oh okay that was the answer for problem 3, but what about #2 o_o?

- anonymous

No, I'm saying that your choice of quadrant 3 is wrong, your answer should be quadrant 2 and 1 other quadrant, think about it

- kittiwitti1

I was looking at my notes... did the teacher teach us the wrong things D:

- kittiwitti1

Is it Q1-Q3?

- freckles

cosine and cotangent have the same sign
that means they could both be positive or they could both be negative
if cosine and cotangent are both negative then what is sine going to be?
if cosine and cotangent are both positive then what is sine going to be?

- kittiwitti1

So Q1-Q3?

- kittiwitti1

Okay

- anonymous

Q3 is not the correct option as I said earlier

- kittiwitti1

Wait what?

- anonymous

in Q3 cos is negative while cot is positive not the same sign at all

- kittiwitti1

My notes say Q3 has both negative values for cos and cot.

- freckles

cosine and cotangent have the same sign
that means they could both be positive or they could both be negative
if cosine and cotangent are both negative then what is sine going to be?
then sine is positive
if cosine and cotangent are both positive then what is sine going to be?
then sine is positive
yes he is right!

- freckles

@Nishant_Garg is right

- freckles

I should have answered my own questions

- kittiwitti1

I just realized I wrote a wrong trig function in my notes...

- kittiwitti1

the* wrong
Nevermind. What about problem B?

- anonymous

you've written wrong dear. Take it from me, never refer to notes for absolute answers because we students tend to write notes during a class while trying to fight off sleep, not very accurate for your notes huh? :P

- kittiwitti1

Lol but I already noticed before you'd said that.
Actually the teacher had miswritten. I should have read the book... forgot the teacher always makes mistakes during lectures (she took 8 minutes till she realized she was going over the wrong problem once)

- kittiwitti1

Even though we tried to tell her five times...
What about problem B? http://prntscr.com/8dia9h

- anonymous

Ouch, for problem 2 you can do the following
remember the unit circle?
we have the following equations
\[x=r\cos(\theta), y=r\sin(\theta)\]
Dividing these we get
\[\frac{y}{x}=\frac{r\sin(\theta)}{r\cos(\theta)}=\tan(\theta)\]
\[\therefore \tan(\theta)=\frac{y}{x}=\frac{7.05}{9.37}\]
Now that you've found 1 trignometric ratio, you can find any other using identities
\[1+\tan^2(\theta)=\sec^2(\theta) \implies \sec(\theta)=\sqrt{1+\tan^2(\theta)} \implies \cos(\theta)=\frac{1}{\sqrt{1+\tan^2(\theta)}}\]

- kittiwitti1

http://prntscr.com/8dilgn
Actually I need help on this first

- kittiwitti1

LOL sorry

- kittiwitti1

Mostly because idk what that symbol means

- anonymous

So which one first?

- kittiwitti1

The one posted just now

- kittiwitti1

I only need to know what the symbol is really

- anonymous

ok
\[\cos(\theta)=\frac{2\sqrt{13}}{13}\]
One ratio you can find right off the start is
secant
\[\sec(\theta)=\frac{1}{\cos(\theta)}=\frac{13}{2\sqrt{13}}\]

- anonymous

\[\in\]
This symbol?

- kittiwitti1

Yes, that one. D:

- anonymous

It means "belongs to"
So what they're saying is theta belongs to quadrant IV
You'll learn more about the symbol as you'll read set theory

- anonymous

what grade are you in?

- kittiwitti1

1st? o_o
(College)

- anonymous

The reason they've given this information will be clear after you read this
suppose we want to find sin(theta) now
we can do
\[\sin^2(\theta)+\cos^2(\theta)=1\]\[\sin^2(\theta)=1-\cos^2(\theta)\]\[\implies |\sin(\theta)|=\sqrt{1-\cos^2(\theta)}\]\[\sin(\theta)=\pm\sqrt{1-\cos^2(\theta)}\]
Since theta is in 4th quadrant, sign will be negative for sin(Theta) so we have
\[\sin(\theta)=-\sqrt{1-\cos^2(\theta)}\]

- anonymous

How come you're studying such basic trigonometry in college? O.o

- kittiwitti1

Didn't take trig in high school.
...the timer reset at 11:59 and erased half the questions so I got 51%... WTF internet

- anonymous

Oh, our syllabus is fixed for school, every student that studies mathematics in high school has studied trigonometry and basic calculus here

- kittiwitti1

http://prntscr.com/8dipej
...I hate this website. Hence why I prefer PHYSICAL tests!

- anonymous

>_<

- kittiwitti1

Well since I didn't finish on time apparently (aka stupid website cleared off half my answers) I guess I'll just close the question = =;

- anonymous

sad :/

- kittiwitti1

Yeah :(
oh well.

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