<p>The function f(x) = rad (3) cos x + sin x has an amplitude of
a) rad 3
b) rad (3) + 1
c) 2
d) 2 rad (3)
e) (rad(3) + 1)/2
Answer is c but i have no clue how to get it</p>
<p>EDIT, nvm, previous post is wrong.</p>
<p>You'd have to add the graph of sqrt(3) cos x to the graph of sin x and find where the graphs are the most constructive - around 45 degrees:</p>
<p>Degrees sqrt(3) cosx sinx Sum
0 sqrt (3) 0 1.732
45 1.22 .707 1.93
90 0 1 1</p>
<p>etc..... so the amplitude is almost 2 ---> the highest sum of the two graphs</p>
<p>Ahh, is there a formula or technique for finding the exact amplitude of such an equation?</p>
<p>On a side note, I plugged in the equation on my graphing calc and used that to find the max y value.</p>
<p>If you put the function in the form acos(x) + bsin(x), the formula for the amplitude is A = sqrt(a^2+b^2), IIRC.</p>
<p>Can't seem to remember the proof though...grr it was so simple when I learned it too...</p>
<p>can't you just graph this?</p>
<p>graph it with a graphing calc, that is.</p>
<p>Yeah, that seems to be the simplest solution.</p>
<p>However, I would like to see a proof for the SQRT(a^2 + b^2) formula.</p>
<p>I think it goes something like this. Let A be the amplitude, x and y are angles:</p>
<p>Acos(x+y)=A(cosx)(cosy)-A(sinx)(siny)
a=A(cosy), b=-A(siny)
a^2=(A(cosy))^2, b^2=(-A(siny))^2=(A(siny))^2
a^2+b^2=A^2[(cosy)^2+(siny)^2]=A^2
A=sqrt(a^2+b^2)
QED</p>
<p>Actually, no...not sure right now how to relate the function and this clearly (let me sleep on it and maybe I'll remember), but this is the gist of it.</p>
<p>If f(x) = a cos(x) + b sin(x)
then f'(x) = -a sin(x) + b cos(x), set = 0 to get max/min</p>
<p>from which a sin(x) = b cos(x) or sin(x) = (b/a) cos(x)
sin^2(x) = (b/a)^2 cos^2(x)
1 - cos^2(x) = (b/a)^2 cos^2(x)
a^2 - a^2 cos^2(x) = b^2 cos^2(x)
cos^2(x) = a^2 / (a^2 + b^2)</p>
<p>At this point, f(x) = a cos(x) + b sin(x) = a cos(x) + (b^2/a) cos(x)
= cos(x) (a^2 + b^2)/a
= sqrt(a^2 / (a^2 + b^2)) (a^2 + b^2)/a
= sqrt( a^2 + b^2)</p>
<p>Seems, questions are reincarnating too...</p>
<p>It's doable without calculus....see what I wrote. I believe the proof is actually doing it backwards--you start out with Acos(x+y)=Acosxcosy-Asinxsiny and show that you can set Acosy and -Asiny as values that happen to be the coefficients of some function f(x)=acos(x)+bsin(x). From there the formula is derived, at least thats how I learned it. Though it is 115am and my brain may not be functioning as well as it should :/</p>