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kittiwitti1
 one year ago
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
kittiwitti1
 one year ago
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw: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
 one year ago
Best ResponseYou've already chosen the best response.1Which problem are you doing? (Sorry I'm multitasking) e_e

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do you have a figure?Talking about problem 1

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.1I'm only wondering where I went wrong, I don't need the whole process explained e_e;;;

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.1Besides they said the sin was correct... My issue is how the tan/cot are incorrect.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.1I don't see what you mean by x... yes I looked at the figure. x_x;

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You did something like dw:1441606633005:dw

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.1Yes but aren't the trig values the same?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Nope!! It makes a slight difference

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.1It does? *trig function values

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Refer 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
 one year ago
Best ResponseYou've already chosen the best response.0As you can see sign changes for sine and cosine but remains same for tangent

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let's find your tan theta, since it's seem to be giving you problem \[\tan(\theta)=\tan(x)=\frac{12a}{5a}=\frac{12}{5}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Actually you've mixed your adjacent and opposite for tangent

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that's why you're getting wrong answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and if your tangent is wrong then obviously cotangent will also be wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0notice you answers you've done \[\frac{5}{12}\] when it should be \[\frac{12}{5}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@kittiwitti1 you there ??

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.1Sorry, working on other problems... What'd I miss?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for problem 1 actually you've exchanged your adjacent and opposite accidentally

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.1facedesks Dx I feel dumb. Okay, thanks; what about the other two?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0problem 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!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441607496705:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[\sin(\theta)=\frac{y}{r} \text{ and } \cos(\theta)=\frac{x}{r} \\ \text{ and you have } r=\sqrt{x^2+y^2} \\ \text{ you can use this \to help you evaluate } \sin(\theta) \text{ and } \cos(\theta)\]

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.1Oh... I seem to have attracted quite the audience... sorry for previous "masstags" ^^;

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441607639710:dw

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.1Wait what are you going on about lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0After>1st quadrant School>2nd quadrant To>3rd quadrant College>4th quadrant

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's an easy way to remember

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.1Yes 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
 one year ago
Best ResponseYou've already chosen the best response.0In problem 3 quadrant is wrong look at it again

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I told you about that just in case you have any trouble remembering what is positive in which quadrant

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.1Oh okay that was the answer for problem 3, but what about #2 o_o?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, 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
 one year ago
Best ResponseYou've already chosen the best response.1I was looking at my notes... did the teacher teach us the wrong things D:

freckles
 one year ago
Best ResponseYou've already chosen the best response.0cosine 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Q3 is not the correct option as I said earlier

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in Q3 cos is negative while cot is positive not the same sign at all

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.1My notes say Q3 has both negative values for cos and cot.

freckles
 one year ago
Best ResponseYou've already chosen the best response.0cosine 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
 one year ago
Best ResponseYou've already chosen the best response.0@Nishant_Garg is right

freckles
 one year ago
Best ResponseYou've already chosen the best response.0I should have answered my own questions

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.1I just realized I wrote a wrong trig function in my notes...

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.1the* wrong Nevermind. What about problem B?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you'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
 one year ago
Best ResponseYou've already chosen the best response.1Lol 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
 one year ago
Best ResponseYou've already chosen the best response.1Even though we tried to tell her five times... What about problem B? http://prntscr.com/8dia9h

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ouch, 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
 one year ago
Best ResponseYou've already chosen the best response.1http://prntscr.com/8dilgn Actually I need help on this first

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.1Mostly because idk what that symbol means

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.1The one posted just now

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.1I only need to know what the symbol is really

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok \[\cos(\theta)=\frac{2\sqrt{13}}{13}\] One ratio you can find right off the start is secant \[\sec(\theta)=\frac{1}{\cos(\theta)}=\frac{13}{2\sqrt{13}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\in\] This symbol?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It 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
 one year ago
Best ResponseYou've already chosen the best response.0what grade are you in?

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.11st? o_o (College)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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
 one year ago
Best ResponseYou've already chosen the best response.0How come you're studying such basic trigonometry in college? O.o

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.1Didn't take trig in high school. ...the timer reset at 11:59 and erased half the questions so I got 51%... WTF internet

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, our syllabus is fixed for school, every student that studies mathematics in high school has studied trigonometry and basic calculus here

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.1http://prntscr.com/8dipej ...I hate this website. Hence why I prefer PHYSICAL tests!

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.1Well since I didn't finish on time apparently (aka stupid website cleared off half my answers) I guess I'll just close the question = =;
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