kittiwitti1
  • kittiwitti1
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • 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
  • kittiwitti1
Which problem are you doing? (Sorry I'm multitasking) e_e
anonymous
  • anonymous
Do you have a figure?Talking about problem 1

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kittiwitti1
  • kittiwitti1
I'm only wondering where I went wrong, I don't need the whole process explained e_e;;;
kittiwitti1
  • kittiwitti1
Besides they said the sin was correct... My issue is how the tan/cot are incorrect.
anonymous
  • 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
  • kittiwitti1
I don't see what you mean by x... yes I looked at the figure. x_x;
anonymous
  • anonymous
You did something like |dw:1441606633005:dw|
kittiwitti1
  • kittiwitti1
Yes but aren't the trig values the same?
anonymous
  • anonymous
Nope!! It makes a slight difference
kittiwitti1
  • kittiwitti1
It does? *trig function values
anonymous
  • 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
  • anonymous
As you can see sign changes for sine and cosine but remains same for tangent
anonymous
  • 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
  • kittiwitti1
Er. Yes...
anonymous
  • anonymous
Actually you've mixed your adjacent and opposite for tangent
anonymous
  • anonymous
that's why you're getting wrong answer
anonymous
  • anonymous
and if your tangent is wrong then obviously cotangent will also be wrong
anonymous
  • anonymous
notice you answers you've done \[\frac{5}{12}\] when it should be \[\frac{12}{5}\]
anonymous
  • anonymous
@kittiwitti1 you there ??
kittiwitti1
  • kittiwitti1
Sorry, working on other problems... What'd I miss?
anonymous
  • anonymous
for problem 1 actually you've exchanged your adjacent and opposite accidentally
kittiwitti1
  • kittiwitti1
-facedesks- Dx I feel dumb. Okay, thanks; what about the other two?
anonymous
  • 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
  • kittiwitti1
B: http://prntscr.com/8dia9h C: http://prntscr.com/8diaew
anonymous
  • anonymous
|dw:1441607496705:dw|
freckles
  • 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
  • kittiwitti1
Oh... I seem to have attracted quite the audience... sorry for previous "mass-tags" ^^;
anonymous
  • anonymous
|dw:1441607639710:dw|
kittiwitti1
  • kittiwitti1
Wait what are you going on about lol
anonymous
  • anonymous
After->1st quadrant School->2nd quadrant To->3rd quadrant College->4th quadrant
anonymous
  • anonymous
It's an easy way to remember
anonymous
  • anonymous
like soh cah toa
kittiwitti1
  • 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
  • anonymous
In problem 3 quadrant is wrong look at it again
anonymous
  • anonymous
3rd quadrant
anonymous
  • anonymous
is wrong
anonymous
  • anonymous
I told you about that just in case you have any trouble remembering what is positive in which quadrant
kittiwitti1
  • kittiwitti1
Oh okay that was the answer for problem 3, but what about #2 o_o?
anonymous
  • 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
  • kittiwitti1
I was looking at my notes... did the teacher teach us the wrong things D:
kittiwitti1
  • kittiwitti1
Is it Q1-Q3?
freckles
  • 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
  • kittiwitti1
So Q1-Q3?
kittiwitti1
  • kittiwitti1
Okay
anonymous
  • anonymous
Q3 is not the correct option as I said earlier
kittiwitti1
  • kittiwitti1
Wait what?
anonymous
  • anonymous
in Q3 cos is negative while cot is positive not the same sign at all
kittiwitti1
  • kittiwitti1
My notes say Q3 has both negative values for cos and cot.
freckles
  • 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
  • freckles
@Nishant_Garg is right
freckles
  • freckles
I should have answered my own questions
kittiwitti1
  • kittiwitti1
I just realized I wrote a wrong trig function in my notes...
kittiwitti1
  • kittiwitti1
the* wrong Nevermind. What about problem B?
anonymous
  • 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
  • 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
  • kittiwitti1
Even though we tried to tell her five times... What about problem B? http://prntscr.com/8dia9h
anonymous
  • 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
  • kittiwitti1
http://prntscr.com/8dilgn Actually I need help on this first
kittiwitti1
  • kittiwitti1
LOL sorry
kittiwitti1
  • kittiwitti1
Mostly because idk what that symbol means
anonymous
  • anonymous
So which one first?
kittiwitti1
  • kittiwitti1
The one posted just now
kittiwitti1
  • kittiwitti1
I only need to know what the symbol is really
anonymous
  • 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
  • anonymous
\[\in\] This symbol?
kittiwitti1
  • kittiwitti1
Yes, that one. D:
anonymous
  • 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
  • anonymous
what grade are you in?
kittiwitti1
  • kittiwitti1
1st? o_o (College)
anonymous
  • 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
  • anonymous
How come you're studying such basic trigonometry in college? O.o
kittiwitti1
  • 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
  • anonymous
Oh, our syllabus is fixed for school, every student that studies mathematics in high school has studied trigonometry and basic calculus here
kittiwitti1
  • kittiwitti1
http://prntscr.com/8dipej ...I hate this website. Hence why I prefer PHYSICAL tests!
anonymous
  • anonymous
>_<
kittiwitti1
  • 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
  • anonymous
sad :/
kittiwitti1
  • kittiwitti1
Yeah :( oh well.

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