Question

\[\frac{ \tan 4 \theta + \tan 2 \theta}{ 1 - \tan 4 \theta \tan 2 \theta}\]

@terenzreignz

Answer 1

I've never worked a problem like this out in my life btw...

Answer 2

<chuckles.

Answer 3

Oh you. :3
Want to play a little game?

Answer 4

Or more specifically, work a little magic? ^^

Answer 5

XD omg... scaring me... sounding like Jigsaw and what not .-. But magic's fine... as long as it had nothing to do with (uk accent) dabbling in the dark arts

Answer 6

has* but I'm ready..

Answer 7

uk accent?
You can actually *hear* me? D:

Answer 8

OMGEEEEEEEEZ YOU HAVE ONE?!

Answer 9

Studied in England.
More to the point... let's start a little deriving game.

It might be a little complicated / daunting, but I'll colour in the parts that I change, at every step, so that you can follow. Are you ready?

Answer 10

Interesting....
I used to live there for a couple of years... I miss it. However I can only immitate lines from Harry Potter with the accent lol >.<
I'll stop getting off topic lol

I'm readyyyy

Answer 11

How many years? Just curious.

And lol let us begin. First, let's start with these identities, these will be crucial.... (I'm being more general here, ok?)
\[\Large \sin(A+B) = \sin(A)\cos(B) + \cos(A)\sin(B)\]\[\Large \cos(A+B) = \cos(A)\cos(B)-\sin(A)\sin(B)\]
\[\Large \tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)}\]

Remember these, and then we'll be good to go.

Answer 12

Around 3... 4 at the most but I was in elementary school though lol
I was just there because the AF stationed my mother over there.
And you?

Understood... I suppose.

Answer 13

Me? About the same. Except the difference between us, probably, is when I first arrived there, I didn't know how to talk yet :P

Ok, are you sure you're ready? :D

Answer 14

Aweeeee cute lol do you have memories o.o

I think so... one question though... what's that symbol with the tangent? o.o

Answer 15

alpha. Another greek letter.
what of it? :P

Answer 16

what does it mean/do

Answer 17

It's just a variable, sheesh :3
normally, 'normal' letters (by this, I mean non-greek ones) represent numbers while greek ones represent angles.

It's not a hard rule, I just wanted some variety XD

Answer 18

Oooooh I see

Ok I'm good now lol

Answer 19

But since it bothers you so much :/
\[\Large \tan(A) = \frac{\sin(A)}{\cos(A)}\]
XD

Answer 20

Okay, ready now?

Answer 21

gracias! I've just never seen it before o.o it might help me adjust if you used them when applicable c:

Readyyyy

Answer 22

Okay, let's begin with this:
\[\Large \frac{\tan(\alpha) + \tan(\beta)}{1-\tan(\alpha)\tan(\beta)}\]
This looks eerily similar to your problem, no?

Answer 23

And before you even ask, \(\beta\) is the Greek letter beta. :P

Answer 24

Got it... Oh goodness... it's a good thing I'm learning about this before entering College xD

Answer 25

LOL college :3

Anyway, using this fact:
\[\large\tan( A) = \frac{\sin(A)}{\cos(A)}\]
The expression becomes:
\[\LARGE \frac{\frac{\sin(\alpha)}{\cos(\alpha)}+\frac{\sin(\beta)}{\cos(\beta)}}{1-\frac{\sin(\alpha)}{\cos(\alpha)}\cdot \frac{\sin(\beta)}{\cos(\beta}}\]

Am I right?

Answer 26

Forgot one detail..
\[\LARGE \frac{\frac{\sin(\alpha)}{\cos(\alpha)}+\frac{\sin(\beta)}{\cos(\beta)}}{1-\frac{\sin(\alpha)}{\cos(\alpha)}\cdot \frac{\sin(\beta)}{\cos(\beta\color{red})}}\]

Answer 27

I'm assuming parentheses aren't to be toyed with lol

Answer 28

Just remember that sine over cosine is just tangent.

Answer 29

But so far so good?
I just changed all the tangents into sines over cosines.

Answer 30

I've got it...

Answer 31

Okay, good.
Now, a tricky step: (new things are in colour)
\[\LARGE \frac{\frac{\sin(\alpha)}{\cos(\alpha)}\cdot\color{blue}{\frac{\cos(\beta)}{\cos(\beta)}}+\color{red}{\frac{\cos(\alpha)}{\cos(\alpha)}}\cdot\frac{\sin(\beta)}{\cos(\beta)}}{1-\frac{\sin(\alpha)}{\cos(\alpha)}\cdot \frac{\sin(\beta)}{\cos(\beta)}}\]

See what I did there?
The coloured fractions are just fancy ways of writing 1, so it doesn't change the expression... right?

Answer 32

Fancy ways of writing 1..?

Answer 33

yeah...
If a fraction has the same numerator and denominator, then it's just basically... 1, right?
\[\large \frac55=\frac44=\frac\pi\pi=1\]

Answer 34

Ooooooh... understood.

Answer 35

Okay, let's simplify those fraction multiplications.... by combining them into just one fraction bar...
\[\LARGE \frac{\color{blue}{\frac{\sin(\alpha)\cos(\beta)}{\cos(\beta) \cos(\beta)}}+\color{red}{\frac{\cos(\alpha)\sin(\beta)}{\cos(\alpha)\cos(\beta)}}}{1-\color{green}{\frac{\sin(\alpha)\sin(\beta)}{\cos(\alpha)\cos(\beta)}}}\]

Answer 36

Still with me?

Answer 37

Before I copy this down... there aren't any typo's are there? xD

Answer 38

There aren't. ^^

Answer 39

Lol thank you ;b I got you..

Answer 40

Also, I'm running out of colours... care to suggest a new one? (aside from blue, red, and green, of course) :P

Answer 41

Is there pink or purple? /.\

Answer 42

hmm...
All right... another sort-of tricky step... replace the 1 with a fancier and more useful 'version' of 1...
\[\LARGE \frac{\color{blue}{\frac{\sin(\alpha)\cos(\beta)}{\cos(\beta) \cos(\beta)}}+\color{red}{\frac{\cos(\alpha)\sin(\beta)}{\cos(\alpha)\cos(\beta)}}}{\color{violet}{{\frac{\cos(\alpha)\cos(\beta)}{\cos(\alpha)\cos(\beta)}}}-\color{green}{\frac{\sin(\alpha)\sin(\beta)}{\cos(\alpha)\cos(\beta)}}}\]

Answer 43

Still with me? I replaced the 1 (in clever fashion) with a fraction having the same numerator and denominator.

Answer 44

I think so...

Answer 45

Hey, don't be pressured... if you even have the slightest doubt or misunderstanding, let me know ^_^

Answer 46

Lol I'll keep that in mind... thank you.

Answer 47

So, no doubts, then?
We can factor something out here:
\[\LARGE \frac{\color{brown}{\frac1{\cos(\alpha)\cos(\beta)}}\left[\color{blue}{\sin(\alpha)\cos(\beta)} + \color{red}{\cos(\alpha)\sin(\beta)}\right] }{\color{brown}{\frac1{\cos(\alpha)\cos(\beta)}}\left[\color{violet}{\cos(\alpha)\cos(\beta)}-\color{green}{\sin(\alpha)\sin(\beta)} \right]}\]

Catching me so far?
The magic is almost complete ^^

Answer 48

Now that I see it, there WAS a typo...
A thousand apologies... the step before this one SHOULD have been...
\[\LARGE \frac{\color{blue}{\frac{\sin(\alpha)\cos(\beta)}{\cos(\alpha ) \cos(\beta)}}+\color{red}{\frac{\cos(\alpha)\sin(\beta)}{\cos(\alpha)\cos(\beta)}}}{\color{violet}{{\frac{\cos(\alpha)\cos(\beta)}{\cos(\alpha)\cos(\beta)}}}-\color{green}{\frac{\sin(\alpha)\sin(\beta)}{\cos(\alpha)\cos(\beta)}}}\]

Answer 49

This is the correction:
\[\LARGE \frac{\color{blue}{\frac{\sin(\alpha)\cos(\beta)}{\cos(\color{black}{\boxed \alpha}) \cos(\beta)}}+\color{red}{\frac{\cos(\alpha)\sin(\beta)}{\cos(\alpha)\cos(\beta)}}}{\color{violet}{{\frac{\cos(\alpha)\cos(\beta)}{\cos(\alpha)\cos(\beta)}}}-\color{green}{\frac{\sin(\alpha)\sin(\beta)}{\cos(\alpha)\cos(\beta)}}}\]

Answer 50

So... that carelessness aside... still with me?

Answer 51

Recap:
This:\[\LARGE \frac{\color{blue}{\frac{\sin(\alpha)\cos(\beta)}{\cos(\alpha ) \cos(\beta)}}+\color{red}{\frac{\cos(\alpha)\sin(\beta)}{\cos(\alpha)\cos(\beta)}}}{\color{violet}{{\frac{\cos(\alpha)\cos(\beta)}{\cos(\alpha)\cos(\beta)}}}-\color{green}{\frac{\sin(\alpha)\sin(\beta)}{\cos(\alpha)\cos(\beta)}}}\]

Becomes THIS:\[\LARGE \frac{\color{brown}{\frac1{\cos(\alpha)\cos(\beta)}}\left[\color{blue}{\sin(\alpha)\cos(\beta)} + \color{red}{\cos(\alpha)\sin(\beta)}\right] }{\color{brown}{\frac1{\cos(\alpha)\cos(\beta)}}\left[\color{violet}{\cos(\alpha)\cos(\beta)}-\color{green}{\sin(\alpha)\sin(\beta)} \right]}\]
after factoring out ^^

Answer 52

Lol it's straight... I got it

Answer 53

Now, this part cancels out, right?
\[\LARGE \frac{\cancel {\frac1{\cos(\alpha)\cos(\beta)}}\left[\color{blue}{\sin(\alpha)\cos(\beta)} + \color{red}{\cos(\alpha)\sin(\beta)}\right] }{\cancel{\frac1{\cos(\alpha)\cos(\beta)}}\left[\color{violet}{\cos(\alpha)\cos(\beta)}-\color{green}{\sin(\alpha)\sin(\beta)} \right]}\]

Answer 54

omgeez so many steps >.<
But that's very magical lol

Answer 55

Still with me? ^^

Answer 56

Yes..

Answer 57

We're left with \[\Large \frac{\color{blue}{\sin(\alpha)\cos(\beta)}+\color{red}{\cos(\alpha)\sin(\beta)}}{\color{violet}{\cos(\alpha)\cos(\beta)}-\color{green}{\sin(\alpha)\sin(\beta)}}\]
right?
We're almost there... does the numerator look familiar to you?
Look back at the three equations I asked you to remember before we started this...

Answer 58

it looks exactly like it...

Answer 59

the denominator, too

Answer 60

Yes it does... so can you simplify the numerator and the denominator? ^^

Answer 61

simplify.. make look simpler...

Answer 62

I think so... since \[\tan(\alpha) = \frac{ \sin(\alpha) }{ \cos(\alpha) }\]

Answer 63

Try not to get ahead of yourself, though.... step-by-step ^_^

Answer 64

o.o
I think it looks like this
idk...
\[\frac{ \sin(\alpha) + \cos(\beta) }{ \cos(\alpha) - \sin(\beta) }\]

Answer 65

No... close, but no.
What is sin(A)cos(B) + cos(A)sin(B) equal to?
Look back to the equations...

Answer 66

sin(A+B)

Answer 67

So... care to revise your simplification? :P

Answer 68

I'm looking back and forth and I don't see where the error is >.<
Why isn't it supposed to look like that?

Answer 69

Well, for starters, as you said,
sin(A)cos(B) + cos(A)sin(B) = sin(A+B)

NOT
sin(A) + cos(B)

:P

Answer 70

hold on hold on sorries gimmie 5 minutes >.<

Answer 71

Sorry... back...
and ooooh
I see...
But
I don't see how I'd get that though

Answer 72

You did. Otherwise, where did THESE come from?

Answer 73

So please, redo the simplification of
\[\Large \frac{\color{blue}{\sin(\alpha)\cos(\beta)}+\color{red}{\cos(\alpha)\sin(\beta)}}{\color{violet}{\cos(\alpha)\cos(\beta)}-\color{green}{\sin(\alpha)\sin(\beta)}}\]

... CORRECTLY this time... :P

PLEASE refer to the equations I gave you initially.

Answer 74

@AngelCriner
Stuck? I'm getting sleepy -_-

Answer 75

A little bit e.e
It's okay though I don't wanna keep you lol
I'll probably figure it out eventually >.<

Answer 76

You will... but you need to focus.
Let me run that by you again...
\[\Large \frac{\color{blue}{\sin(\alpha)\cos(\beta)}+\color{red}{\cos(\alpha)\sin(\beta)}}{\color{violet}{\cos(\alpha)\cos(\beta)}-\color{green}{\sin(\alpha)\sin(\beta)}}\]

What is the numerator here equal to?

Answer 77

I'm just not seeing how I can move anything around with that >.< I looked at the equations and how you simplified them... but I'm stuck. Are you saying it's supposed be simplified down to \[\frac{ \sin(\alpha + \beta) }{ \cos( \alpha + \beta}\] *with a parenthesis after Beta* ? Because I don't see how I can do that...

Answer 78

unless

Answer 79

Of course it is.

Answer 80

cos and cos on the rhs cross each other out and the sin and sin cross each other out on the lhs leaving sin(A+B) + cos(A+B) / cos(A+B) - sin(A+B)

Answer 81

It's that simple.
sin(A)cos(B) + cos(A)sin(B) = sin(A+B)
cos(A)cos(B) - sin(A)sin(B) = cos(A+B)

So why SHOULDN'T
\[\Large \frac{\color{blue}{\sin(\alpha)\cos(\beta)}+\color{red}{\cos(\alpha)\sin(\beta)}}{\color{violet}{\cos(\alpha)\cos(\beta)}-\color{green}{\sin(\alpha)\sin(\beta)}}\]
be equal to\[\huge\frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)}\]?

Answer 82

Wait, what?!
WHAT crosses each other out?

Answer 83

No, please, continue.

Answer 84

convinced yet?

Answer 85

Ugghhhhh I'm dying...

that's all I'm seeing and I know for a fact it's incorrect

how does the denominator even turn into addition like the subtraction sign is totally obliterated.. it's just gone e.e where do the other a's and B's go

I'm sorry my mind is really not grasping this concept