<p>Hai guys,
Here's a thread to ask questions, post problems, and remember those pesky formulas.
I'll start off with the dot product!
Dot product of vector A and vector B = A1B1+ A2B2
Also, if the dot product is 0, then the two vectors are perpendicular!!!</p>
<p>Period of a sine or cosine curve in the form y=sin(ax) or y=cosine(ax) is (2pi)\a . </p>
<p>Period of a tan curve in the form y=tan(ax) is (pi)\a,</p>
<p>Awesome! Also:
*Combinations = order doesn’t matter
- Permutations = order DOES matter</p>
<p>howd u guys study</p>
<p>vector product (cross product):
| i j k |
| a1 a2 a3 |
| b1 b2 b3 |</p>
<p>(it’s supposed to be a determinant, but the formatting messed it up) </p>
<p>This is a good idea for a thread.</p>
<p>This is a question from my Barrons book and I’m confused why the correct answer is (A)</p>
<p>What is the range of the function f(x)=5-6sin(pix+1)?
(A) <a href=“B”>-6,6</a> <a href=“C”>-5,5</a> <a href=“D”>-1,1</a> <a href=“E”>-1,11</a> [-11,1]</p>
<p>I think the answer should be (D); oftentimes Barron’s makes mistakes:</p>
<p>Max of sin is 1 and min is -1, so least f (x) can be is 5-6=-1 and greatest is 5-(-6)=11</p>
<p>@angehere- i remember that question and thinking it was weird… i just went with the flow ;)</p>
<p>sin2x= 2sinxcosx
cos2x= cos^2(x)-sin^2(x) = 2cos(x)^2-1 = 1-2sin^2(x)</p>
<p>And how do you find the range of a function without typing it in your graphing calc?</p>
<p>Wouldn’t it be easier to just your calculator?</p>
<p>@angehere, it’s A. because the range of the function f(x)=sin(x) is -1,1. If you look closely, the center of this graph has been shifted and there is a -6 in front of the sin(x) part. This means that the graph is reflected across the x axis and is stretched by a factor of x6. So the new range would consequently be -6,6.</p>
<p>Sorry if that was confusing…</p>
<p>but then its vertically shifted up 5 units retart</p>
<p>cos(a+b)=cos(a)cos(b) - sin(a)sin(b)
cos(a-b)=cos(a)cos(b) + sin(a)sin(b)
sin(a+b)=sin(a)cos(b) + cos(a)sin(b)
sin(a-b)=sin(a)cos(b) - cos(a)sin(b)</p>
<p>Wait do we need to know sum and difference identities because that was not there in barons… better question, wat trig identities shud I memorize? </p>
<p>Sent from my Desire HD using CC App</p>
<p>Upon consulting with wolfram, I confirm that (d) IS the answer</p>
<p>Just thought of a fantastic way to remamber combinations/permutations-
COD, or call of duty, in that combinations order doesn’t (matter). YAY.
Remember, when solving for an absolute value of an imaginary number, plot it on the imaginary plane, then find the hypotenuse.</p>
<p>Awesome! I also believe that the modulus of an imaginary number is a^2 + b^2 ?
Also, remember that the diagonal of a cube is sqr root of (l^2+w^2+h^2)
Any other quirky concepts?</p>
<p>@angehere</p>
<p>i just used my graphing calculator and graphed it. answer is d</p>
<p>I still think it’s A. Here’s why…this is a graph of y=sin(x) that is
- reflected vertically (about the x-axis)
- stretched by a factor of 6
- shrunk by a factor of pi
- shifted to the left 1/pi
- shifted up 5</p>
<p>As you may recall, the stretching, shrinking, and reflecting are ALWAYS done first before the shifting. So we should have a graph that is reflected across the x-axis, stretched by x6, and shrunk by pi. We can essentially ignore the shrinking/reflecting in this problem. Now that the graph has been stretched by x6 we know the range is -6, 6. AFTER ALL THIS, we SHIFT the graph. It becomes evident that shifting will NOT change the range.</p>
<p>^Except there is a vertical expansion…
Just plug the equation into the graphing calculator. Answer D is most definitely correct.</p>