Question

Anyone who took Pre-Calculus and understands it, please help me. I'm drowning.

Answer 1

Did you have a question that you needed help with?

Answer 2

Try two whole topics that I need help with. But if it needs to be narrowed down to a question, how about this.

Solve the equation on the interval [0,2pi) \[\cos2x=\sqrt{3}/2\]

Answer 3

So what part of that are you having trouble with?

Answer 4

Well, you see, our teacher hasn't covered this or reviewed it since...2nd quarter? And now it's on our final and I don't understand how to do any of the identities or logarithms. So...basically I'm having trouble with the whole equation.

Answer 5

Well, what about the unit circle?

Answer 6

I don't remember how to use the unit circle. Do I use it for this? And if so, how?

Answer 7

Lets start with let \(\theta=2x\) for a quick substitution. That will have to go backwards later, but it is for a reason.

Now, look at the unit circle:
http://en.wikipedia.org/wiki/File:Unit_circle_angles_color.svg

Where does \(\cos\theta = \frac{\sqrt{3}}{2}\) ?

Answer 8

\[\pi/6\] and \[11\pi/6\]

Answer 9

Yes. Now, that was for \(\theta\), not for x. So, we have to convert it into x. Also, your x can be between \([0,2\pi)\), so there may be more answers!

Answer 10

So, if we used \(\theta=2x\), what does that make those first two points in terms of x?

Answer 11

Doubled?

Answer 12

\(\theta=2x\implies \frac{1}{2}\cdot\theta=\frac{1}{2}\cdot2x\)....

Answer 13

Oh, so it would be half?

Answer 14

Yes. So, what is half of those first two points? Those we know will be answers.

Answer 15

Ok. That would be... \[22\pi/6\] and \[2\pi/6\]

Answer 16

No...that's wrong.

Answer 17

You did the 2, not the half. Hehe. Yah. 6 becomes 12...

Answer 18

\[\pi/12\] and \[11\pi/12\]

Answer 19

So put \(\frac{\sqrt{\pi}}{12}\) and \(\frac{\sqrt{11\pi}}{12}\) aside as answers. We know those are between \([0,2\pi)\). The next thing is to see if there are more.

Now, the basis we found was the points \(\frac{\sqrt{\pi}}{6}\) and \(\frac{\sqrt{11\pi}}{6}\). If there were more points, they would be co-terminal angles of those. What would be the next co-terminal angle for each of those? And, is half of the next co-terminal angle \(\ge 2\pi\)?

Answer 20

Aak... I don't know why I put the root in there.. . LOL

Answer 21

Haha

Answer 22

OH, copied the text and pasted... and forgot to remove the root. LOL

Answer 23

lol

Answer 24

Could you remind me what a co-terminal angle is? Isn't it an angle on the opposite side of the unit circle?

Answer 25

So put \(\frac{\pi}{12}\) and \(\frac{11\pi}{12}\) aside as answers. We know those are between \([0,2\pi)\). The next thing is to see if there are more.

Now, the basis we found was the points \(\frac{\pi}{6}\) and \(\frac{11\pi}{6}\). If there were more points, they would be coterminal angles of those. What would be the next coterminal angle for each of those? And, is half of the next coterminal angle \(\ge 2\pi\)?

That is more like it! Ignore the man behind the curtain!

Answer 26

Hahaha! Nice reference.

Answer 27

co means same, terminal means end.

Answer 28

I know what the word means. I'm saying what does it mean in the context of Pre-Calc?

Answer 29

Exactly that. Angles that have the same end point.

Answer 30

Oh. Okay. Haha.

Answer 31

So \(\pm360^\circ\) or \(\pm 2\pi\), with next being the +.

Answer 32

Oh! I remember that! Ok.

Answer 33

Would it be \[23\pi/6\] and \[13\pi/6\]

Answer 34

\[\frac{\pi}{6}+\frac{12\pi}{6}=\frac{13\pi}{6}\implies \frac{13\pi}{12}\]
\[\frac{11\pi}{6}+\frac{12\pi}{6}=\frac{23\pi}{6}\implies \frac{23\pi}{12}\]

Answer 35

Wait, how did you get the denominator to 12?

Answer 36

I am just showing the x version. Remember, we still do the theta to x on these.
Now, the stopping point is \([0,2\pi)\). Either of those over that line?

Answer 37

Ummm...Yes.
Both of them are.

Answer 38

What is \(2\pi\) when done as a fraction over 12?

Answer 39

Ohhh...I'm tired, not thinking straight. Lol sorry.
\[24\pi/12\]

Answer 40

All I did in one step was this stuff:\[\theta =\frac{13\pi}{6}\implies 2x =\frac{13\pi}{6}\implies \frac{1}{2}\cdot 2x =\frac{1}{2}\cdot \frac{13\pi}{6}\implies x =\frac{13\pi}{12}\]
Didn't mean to catch you off guard.

Correct, \(2\pi=\frac{24\pi}{12}\). So, are either of those two answers \(\ge \frac{24\pi}{12}\)

Answer 41

No.

Answer 42

So those are withn the domain and therefore answers.

Answer 43

Now we have 4 answers. So, based on the highest answer, do you think we need to search for more?

Answer 44

No. We have enough. This test is multiple choice, by the way. And one of the answers is \[\pi/12, 11\pi/12, 13\pi/12, 23\pi/12\]

Answer 45

So, we're good! :-)

Answer 46

In formal notation, the answer is:
\[x=\left \{\frac{\pi}{12},\frac{11\pi}{12},\frac{13\pi}{12},\frac{23\pi}{12}\right \}\]which you don't need for multiple choice...

Answer 47

Yeah, lol
Ok. Now the next one looks confusing too, if it's not too much trouble.

Answer 48

If you look at the sequence, the next one would have been \(\frac{25\pi}{12}\). Comes in handy if you need the answers as a middle part of something else. Find the sequence, and then they are easy to see.

So what is the next one?

Answer 49

Thank you so much, by the way. :)

Answer 50

The next one is...

Answer 51

\[\tan 2x-\tan x=0\]

Answer 52

OK, so you need to get that to one angle. Right now you have two angles, 2x and x.

Answer 53

So how would I go about that?

Answer 54

Have a list of trig identities?

Answer 55

Ummm...yes I do. Lemme grab it.

Answer 56

Got it.

Answer 57

We want something that changes \(\tan 2x\) into \(\tan x\) or the other way around.

Answer 58

Well, there's the Double-Angle Formulas.

Answer 59

Yep! And tan is the ugly one, but oh well.

Answer 60

\[\tan 2u=2\tan u/1-\tan ^{2}u\]

Answer 61

Ok, so there goes 2x. Assassinated and replaced with something "friendlier" to manipulation.

Answer 62

Lol good.

Answer 63

So, you have a fraction minus a something. Common denominator time.

Answer 64

I feel so dumb. What do I to get the common denominator? Can I just put the tanx on the denominator of the other one?

Answer 65

You must use 1 in disguise. Remember that for any value, equation, etc. \(\frac{n}{n}=1\)

Answer 66

So, I put the tanx over a 1?

Answer 67

\[\frac{1-\tan ^{2}u}{1-\tan ^{2}u}=1\]And you can multiply anything by 1 and not change an equation.

Answer 68

Ohhh! That makes sense. But there isn't a 1 in the equation.

Answer 69

OOOOO...sorry. I'm not thinking.

Answer 70

\[\tan 2x-\tan x=0\implies \frac{2\tan x}{1-\tan ^{2}}-\tan x=0\]What we need to change is that -tanx by multiplying it by 1.

Answer 71

o ok

Answer 72

So, if we have that special form of 1.... what would happen if we did a little multiplication?

Answer 73

It would change the denominator to the common denominator.

Answer 74

Exactly!

Answer 75

Wow. I way overthink these problems.

Answer 76

So, how do I proceed when I have everything common denominator-y.

Answer 77

\[\frac{2\tan x}{1-\tan ^{2}x}-\tan x\cdot \frac{1-\tan ^{2}x}{1-\tan ^{2}x}=0\implies \frac{2\tan x}{1-\tan ^{2}x}- \frac{(\tan x)(1-\tan ^{2}x)}{1-\tan ^{2}x}=0\]

Answer 78

Well, now that you can see it. What is next?

Answer 79

I want to say...multiply the top of the right.

Answer 80

Distribute the tangent of x.

Answer 81

That looks good. Might have to factor it at some point though.... hmmm... I wonder if there is a factor by grouping hiding there somewhere.

Answer 82

\[\frac{(2\tan x)-(\tan x)(1-\tan ^{2}x)}{1-\tan ^{2}x}\]
\[\frac{\tan x[(2)-(1)(1-\tan ^{2}x)]}{1-\tan ^{2}x}\]
\[\frac{\tan x[2-1+\tan ^{2}x]}{1-\tan ^{2}x}\]
\[\frac{\tan x[1+\tan ^{2}x]}{1-\tan ^{2}x}\]
Herm.

Answer 83

Not quite what I expected... but I think it will work. Multiplying and combinding would have given the same thing.

Answer 84

Ok.

Answer 85

\[\frac{(2\tan x)-(\tan x)(1-\tan ^{2}x)}{1-\tan ^{2}x}\]
\[\frac{2\tan x-\tan x+\tan ^{3}x}{1-\tan ^{2}x}\]
\[\frac{\tan x+\tan ^{3}x}{1-\tan ^{2}x}\]
And you can see how that ends up the same. either way is just as good.

Answer 86

Do we have to do all this just to find the answers? Aren't we going to use the unit circle?

Answer 87

Well, I was hoping to get rid of the bottom!

Answer 88

Ok.

Answer 89

So there is a little more work.... Cause I did not get the factors I expected. See, I was hoing the top would become something that canceled the bottom. =/

Answer 90

I hate my teacher. Why would he give us 129 questions for our final and not be there for questions, and when you ask questions in the short time you DO see him, he says he can't help out?

Answer 91

It's bad teaching.

Answer 92

Fun. OK. So lets see... that bottom does factor. It is a difference of squares. Also, it can be replaced ... hmm... cotangent identitiy I think.

Answer 93

Ah, secent... hmmm... might not be bad.

Answer 94

Look at the Pythagoran identities.

Answer 95

That won't work. The one you're thinking of is \[1+\tan ^{2}\theta=\sec ^{2}\]θ

Answer 96

Yes.... but it might... I am trying something, which is an important thing in trig.

Answer 97

Alrighty.

Answer 98

I found something too ugly to contemplate. LOL! Want to see?