JabbaMatheez - The Touch Up.

APPARENTLY I got 1000 marks for finding 'X' SO, I guess I don't have to come to class anymore. =P

We did a little touch up class before we headed on to the next unit. On top of that, we conversed about various things, like when we should have the math test, (Some crazy guy suggested we do the test THEN the pre-test. What a guy huh?), DJ K's baby, Justus' remix to "Apologize" and many other random things... that unfortunately had nothing to do with math. BUT Hupsha, hupsha, quick like a bunny, we went off to work.

If you look at the second slide, you'll see we did a simple warm up and proved some identities. REMEMBER! When the question asks that you PROVE and I mean PROVE, create your GREAT WALL OF JABBA! Which is to say YOU CANNOT CROSS THAT LINE! But when it says SOLVE then you may cross that line and do whatever you wish. That first one on the left side of the slide was done none other than ELVEN. Whom was remarked as... HAWT. So anyways, he simply changed 1 + tan² x into seceant, which is 1/cos² x. He then multiplied the sin²x which would end up becoming tan² x. And so both sides are the same.

On the right side, ben started off by changing seceant and coseceant into their 1 / sine and 1 / cosine bretheren. On the other side of the Great Wall of Jabba, he changed it to 1 / cos² xsin²x, so that when he finished the left side they would be identical. QED!

Intermission...

And Now Back to our Regular Programming...

Closed Captioning brought to you by... Rence

This Blog is sponsored by DJ K!!!

Now if you look at the 3rd slide, it says solve so what does that mean class!? That's right! Mr. K's buying us donuts for tomorrows class! LOL I wish but actually it means you can cross the Great Wall of Jabba. And that's exactly what Thi does in this slide, marked in GREEN. He moves over the 1 and divides 2 from both sides to get cos²x = 1/2. He then square rooted it BUT he forgot the + /- sign. Don't forget that, write that down. Because of that, he missed two of the solutions and so his solution was not completely wrong, but incomplete. Props to Thi for going up to the board and doing that, bcause chances are we would've made the same mistake. Except Ben, but hey, Ben's a genius.

Marked down in BLACK, DJ K's got the elegant way of answering the question. So this is how he breaks it down, and no I don't mean by doing flares and what not. I mean how he takes the 2cos² x - 1 and breaks it down to (√2cosx + 1)(√2cosx - 1) = 0.

From that we can derive that cosx = 1/√2 and cosx = -1/√2 but WWWAAAIIITTT! We have to rationalize that! So it by multiplying both the top and bottom by √2 we get cosx = √2/2, -√2/2, and that's how DJ K get's the other two solutions. Ya'll dig or what?

Continuing on to slide 4, we have a similar question, but sine and cos are in the same equation(2cos²x = 2 + sinx)!

Oh No! Like DJ K said, we couldn't probably do it weeks ago and say "DJ K, I'm sorry but I only work with one way equations. Throw in Sine and Cosine, sorry, no can do." And he's right, but us, being smarter than the average bear find that cos²x = 1 - sin²x. So now everything's sin²x. Can we do it? Yes we can!

So then we basically get everything off to one side in this equation (is now 2(1 - sin²x) = 2 + sinx --> 2 - 2sin²x = 2 + sinx *since the 2's cancel* --> 0 = 2sin²x + sinx) we can then factor out sinx (sinx(2sinx + 1)) in which we find that sinx = - 1/2.

Well, we find through that, that the solution is 11π/6 + 2 kπ ; KЄ I and 7π/6 + 2kπ ; KЄ I. On the next slide we practically do the same thing. Different trig functions and numbers is all.

Another intermission

Continuing to our next slide, we continue with our regular scheduled programming.

In this slide we are given sinα = 4/5 and cos β = -5 / 13 where cosα <> 0. So to find those, we know that sinβ and cosα are in quadrant 2 using Pythagoras's theorem & a picture of the unit square like so.

5² - 4² = 3² (25 - 16 = 9 --> √9 = 3) and 13² - 5² = 12 ² (169 - 25 = 144 --> √144 = 12)

It then asks us to find what Tan(α - β) is so we basically combine Sin(α - β) /Cos(α - β)

GREAT! So now we have cosα = -3/5 and sinβ = 12 / 13. So Now we can get Tan(α - β) .

I know it's homework, but I'll post up my answer so that others can compare. Correct me if I'm wrong PLEASE!

Sin(α - β) = SinαCosβ - CosαSinβ. Cos(α - β) = CosαCosβ + SinαSinβ.

Then, set it up like this. SinαCosβ - CosαSinβ/CosαCosβ + SinαSinβ.

Replace them. (4/5)(-5/13) - (-3/5)(12/13) / (-3/5)(-5/13) + (4/5)(12/13) --> (-20/65) - (36/65) / (15/65) + (48/65) --> (-56/65) / (63/65)

*MULTIPLY BY THE RECIPROCAL* (-56/65) * (65/63)

* 65's REDUCE* -56 / 63. So I got an answer of Tan(α - β) = -56 / 63

That's all for today folks! Remember, we start a new unit tomorrow!!

P.S. HEY fellow JabbaMatheez! We still have our PICTURES due after the break so don't forget that while you're all having fun and having adventures on SPRING BREAK '08!

Oh wait, there's still the masking of the next Scribe!

I will pass the Jabba Scribe on to... FRANCIS!!

~Rence OUT!