Only math geniuses can answer this question...

Test Preparation
1

<p>Can anyone explian this question?</p>

<p>What is the amplitude of the function Y=a(cos(x))+b(sin(x)).</p>

<p>Does anyone know how good rusen's 15 tests (math iic) are?</p>

2

<p><em>edit</em> lol nm on my answer</p>

3

<p>I think a(root2 +b*root2) since the max occurs at x=Pi/4</p>

4

<p>graph it on your calculator and find out</p>

5

<p>KElly05</p>

<p>WRONG</p>

<p>Keep coming people. Anyone know the answer to my original question?</p>

6

<p>The answer is the square root of a squared plus b squared.
How...</p>

<p>15 realistic practice tests! How good/hard are they?</p>

7

<p>I think it's an optimization problem, find the derivative and find its zeros.
x=arctan(b/a)
Therefore, y=a^2/(a^2+b^2)^1/2 + b^2/(a^2+b^2)^1/2
y=(a^2+b^2)^(1/2)</p>

8

<p>Darn it stix, you posted the answer while I was solving it.</p>

9

<p>Yes I noticed. See that my answer went away? Thanks for making me look like the fool I am. <em>cries</em></p>

10

<p>Oh wait a minute, I used calculus to solve it. This is supposed to be Math IIC?</p>

11

<p>its not really calc; if and only if the period of Asinx and Bcosx are equal, then you can determine the amplitude of Acosx+Bsinx, and it is sqrt(A^2+B^2)</p>

12

<p>Sorry!!</p>

<p>So, how bout them 15 realistic tests.</p>

<p>SOMEONE ANSWER ME!!!!!!!!!!!!!!!!!!!</p>

13

<p>I don't understand how you can arrive at sqrt(a^2+b^2) without using calc to prove it. Is it simply a principle one memorizes? Then calculus is used to derive/proove it, as simple as that.</p>

14

<p>I'm only kidding =)</p>

<p>Would converting sin to cos via the sin(x-90) = cos(x) or whatever that identity is help at all?</p>

15

<p>Barron's test number three question 50. You have to multiply by a bunch of different numbers and use trig identities to simplify. I was wondering if there was an eaisier way to do it...</p>

16

<p>um no, you can prove it without calculus, i remember my precal teacher doing it about 2 years ago</p>

17

<p>Interesting. Show me the proof then.</p>

<p>EDIT: Trig identities, huh? Interestingly enough, if you want a easier way to solve it, calc would be that way. It's really easy with calculus, a couple of equations.</p>

18

<p>Start by multipling by the square root of a squared plus b squared over the square root of a squared plus b squared. I have to go... Full proof tomorrow morning.</p>

19

<p>Oh I see, you just multiply by the hypotenuse of the triangle to convert the sines and cosines to a's and b's. That's pretty much just the full proof then. Ingenius!</p>

20

<p>I still don't get how you figured it out without calculus</p>