Question

Find all angles(theta) between 1 and 2π for which cot(theta) = √3.
How would you do this without a unit circle?

Answer 1

1) Why?
2) sqrt(3) = cot(x) = cos(x)/sin(x) = sqrt(3)/1 = [sqrt(3)/2]/[1/2]
3) If that doesn't look like a 30-60-90 right triangle, I don't know what does!

Answer 2

Where did you get the last values from? for #2 (the √3/2 ÷ 1/2)?

Answer 3

Magic? I happen to know that a 30-60-90 right triangle has sine and cosine equal to 1/3 and sqrt(3)/2, or the other way around. You should know this, too.

Don't forget that this is in Quadrant I. There is another solution in Quadrant III where both values are negative?

Answer 4

Seems like I will need a lot of memorizing to do... Am I supposed to know automatically that this is in quadrant 1 from the positive value?

Answer 5

sine and cosine are positive. Absolutely you should know this is Quadrant I.

Answer 6

So what would be angle values?

Answer 7

It's still a bit confusing..

Answer 8

We established that cos(x) = sqrt(3)/2 and sin(x) = 1/2. That's 30º. You find the Quadrant III solution.

Answer 9

Mmm ok these tangent and cotangent functions can be a little tricky :)

Answer 10

I agree! lol

Answer 11

So like... let's think for just a sec.
Lemme give a quick example.

\[\Large \tan 60^o \quad=\quad \frac{\sin 60^o}{\cos60^o} \quad=\quad \frac{\left(\dfrac{\sqrt3}{2}\right)}{\left(\dfrac{1}{2}\right)}\quad=\quad \sqrt3\]

Answer 12

If we had started with \(\Large \tan\theta=\sqrt3\) can you see how it would be tough to reference it back to sine and cosine since it's simplified down?
That's what makes this problem a little tough :P heh

Answer 13

If you remember your reference angles for tangent and cotangent, it makes this a little easier.

Answer 14

\[\Large \cot \theta=\sqrt3\]
Do you have this reference angle memorized or no? :D

Answer 15

no :/

Answer 16

You and your strawberry ways -_-

Answer 17

hehe :D

Answer 18

Hmmmmm how to explain this then :o hmmm

Answer 19

I wish I had them memorized...

Answer 20

You are very good at explaining... :D

Answer 21

Try to memorize the reference angles for sine, cosine and tangent.

Here are a couple important ones for tangent.\[\Large \tan \theta=\sqrt3 \qquad\to\qquad \theta=\pi/3\]\[\Large \tan \theta=\frac{1}{\sqrt3} \qquad\to\qquad \theta=\pi/6\]

Answer 22

Since we are dealing with cotangent, we can relate it back to tangent using this identity:\[\Large \cot \theta =\frac{1}{\tan \theta}\]

Answer 23

\[\Large \cot \theta=\sqrt3 \qquad\to\qquad \tan \theta=\frac{1}{\sqrt3}\]
Understand what I did there?
Since tangent is the `flip` of cotagent, I did a lil magic :o

Answer 24

ohhh, yess, I do see :D

Answer 25

you are so quick at typing up all that btw ~

Answer 26

lol :3

Answer 27

So as we listed above: \(\Large \tan \theta=\dfrac{1}{\sqrt3} \qquad\to\qquad \theta=\pi/6\)

Answer 28

And this is the same as \(\Large \cot\theta=\sqrt3\).

Answer 29

So our reference angle is \(\Large \theta=\pi/6\).
Or 30 degrees, as you already figured out earlier it seems :)

Answer 30

Hmm so we need the other one?

Answer 31

I think so, the one for quadrant 3??

Answer 32

There are a couple of ways to approach this.
We can use this,\[\Large \cot \theta \quad=\quad \frac{\cos \theta}{\sin \theta}\quad=\quad \sqrt3\]
and then think about where else on the unit circle we're getting sqrt3.
It will be where sine and cosine are both negative, right? That's how we can get a positive, when dividing negative by negative.

Answer 33

There's another approach we can try though.
This might be a tad easier.

Answer 34

Since tangent and cotangent have period \(\Large \pi\), they will go through their full cycle of values in pi and then start to repeat after that.

So we could simply add \(\Large \pi\) to find the other value.

Answer 35

So our first angle was \(\Large \dfrac{\pi}{6}\),

Our next one will be \(\Large \dfrac{\pi}{6}+\pi\)

Answer 36

Too much maf? brain esplode? +_+

Answer 37

lol, yes way to much math for me that this hour... it's like 1:17am here :O but I need to turn in my quiz tomorrow and need to understand all this.. you are such a big help though ! :)

Answer 38

But I am not sure why it would be to add π to π/6..

Answer 39

So think back to sine and cosine.
Let's imagine we're at an angle of pi/6 on the unit circle.
Try to picture it.

If I tell you to add 2pi to that angle. I'm telling you to spin around the circle and last in the same spot, right?

Answer 40

yesss :)

Answer 41

So if sin pi/6 = 1/2

What does sin (pi/6 + 2pi) =

Answer 42

Hint: we landed in the same spot after we spun!! :O
It should still give us 1/2 right?

Answer 43

ohhh, yea you are right

Answer 44

I was overthinking it...

Answer 45

Well the same thing happens with the tangent/cotangent functions.
But they start to repeat when you spin HALF WAY around the circle, not a full spin.

Answer 46

cot (pi/6) = sqrt3

cot (pi/6 + pi) = sqrt3

Answer 47

Is that something I should memorize? :D

Answer 48

The thing that I just wrote with the sqrt3?
Or the thing about the half way spinning? :O

Answer 49

half way spinning :) lol

Answer 50

Yes, try to remember it like this.

Sine and Cosine have a period of 2pi.
(They repeat their values after rotating 2pi more around the unit circle).

Tangent and Cotangent have a period of pi.
(They repeat their values after rotation pi more around the unit circle).

Answer 51

okay, I think this will really help me, yess! :) Thanks, it seems like the teacher does not explain as good as you :D

Answer 52

Well the teacher can't give you personal 1 on 1 time ^^ So it can be hard to get every little detail hehe

Answer 53

sOo very true...

Answer 54

Was that all of the problem ??

Answer 55

I wish you could tutor me in for real.. I would be a pro in trig :D

Answer 56

Ya I guess the way you would want to write your answer would be,\[\Large \theta=\frac{\pi}{6}, \quad\frac{7\pi}{6}\]
Do you understand how I got the second angle?
Did you have any trouble adding the pi to the fraction?

Answer 57

Yes fo sure I understand it with how well you explained it hehe. Thanks SO much!

Answer 58

np \c:/ go to bed!!

Answer 59

:D