<p>I'm making this thread for the people who are in the same boat as i am and prepping for calc in the fall. Each day starting tommorow I will post a few problems. They will start out very simple and will get harder but will only cover the tools needed to do well in ap calc.</p>
<p>That sounds like a great idea...what kind of problems will be posted?</p>
<p>ok the first set of problems... as i said the problems start out very very easy so...</p>
<p>1.) simplify by factoring:</p>
<p>(5-x)/(x^2-25)</p>
<p>Trigonometric Identities:</p>
<p>1.) cos (2x) =
2.) sin (2x) =
3.) tan (2x) =</p>
<p>Once these are solved I will post new problems.</p>
<p>(5-x)/(x^2-25)
= -1*(x-5)/[(x+5)(x-5)]
= -1/(x+5)</p>
<p>cos(2x) = cos^2(x) - sin^2(x)
= 2cos^2(x) - 1
= 1 - 2sin^2(x)</p>
<p>sin(2x) = 2sin(x)cos(x)</p>
<p>tan(2x) = 2tan(x)/[1 - tan^2(x)]</p>
<p>Nooooo! I've forgotten all those. And BC starts in two weeks. Dang trig formulas. And what were those special derivatives? Like dy/dx of sinx at 0=1, and dy=dx of cosx at 1=0 or something like that. And I totally don't know how to integrate anymore. First week back will be fun.</p>
<p>1) -1/(x+5)
ID's
cos2x=cos^2(x) - sin^2(x) or 2cos^2(x) - 1 or 1 - 2sin^2(x)
sin2x=2sinxcosx
Tan2x=2tan(x)/(1 - tan^2(x))</p>
<p>hey do you think smartguy, you could like post the number of days we're on, like this is day 1?</p>
<p>yes that was day 1</p>
<p>ok umm, i've never dealt with trig identities before so if #18 or anyone else could clarify what they did</p>
<p>Those are trig identities - they are just "given," meaning the transformations are derived from formulas.</p>
<p>smartguy, trig identities have no work involved...all you can do is memorize them
I'll see if I can explain how the identities are used....</p>
<p>For example, the three identities we just defined are the "Double-Angle" formulas.
Say you have the sine of angle x. The sine of an angle double that size, angle 2x, would be equal to 2<em>sin(x)</em>cos(x).</p>
<p>hope that helps</p>
<p>ok time for day 2 questions, the questions will get harder on day 3/4</p>
<p>Simplify Each Expression</p>
<p>1.) (1/(x+h)) - (1/x) </p>
<p>2.) (2/x^2)/(10/x^5)</p>
<p>3.) ((1/(3+x)) - (1/3)) / x</p>
<p>4.) (2x/x^2-6x+9) - (1/x+1) - (8/x^2-2x-3)</p>
<p>1) (-h)/(x^2+h)
2) (x^3)/5
3) (-1) / 3x+9
4) (x^2 - 12x + 33) / { ((x-3)^2 )(x+1) }</p>
<p>not sure if the above answers are correct or not..</p>
<p>could someone please post FULL solutions to the day 2 problems as just the answers arent really helping anyone</p>
<p>I apologize. I find it hard typing the solutions.. shall keep that in mind for the the next set though... Anyways, Here are the solutoins for the first two... </p>
<p>1) (1/x+h) - ( 1/x ) = ( x - x - h ) / ( x^2 + xh ) = ( -h / x^2 + xh )
2) (2)/(x^2) . ( x^5 ) / ( 10 ) = x^3 / 5</p>
<h1>3, 4? anyone</h1>
<ol>
<li>(x^2 + 15)/((x + 1)(x^2 - 6x + 9))</li>
</ol>
<p>before i post day 3 i need a full solution for #3</p>
<h1>3 ((2x/3)-1)/(x^2+3x)</h1>
<p>so is day 3 going to be more difficult?</p>
<p>jai i believe ur number 1 for day 2 is wrong so if somebody could post a new solution and zumy yes, day 3 is more difficult</p>
<p>What is wrong with -h/(x^2 + xh) for #1, day 2?</p>