<p>oh (a+b)/2>=root ab
I think that's what he/she used</p>
<p>Yeah... this is one of those problems that is challenging until you learn calc. These kinds of things appear a lot on non calculus math competitions. I would advise learning basic integral and differential just for problems like these. Very useful.</p>
<p>Here is a non calculus solution:</p>
<p>a(cos(x))+b(sin(x))
=sqrt(a^2+b^2)<em>(a/sqrt(a^2+b^2)</em>cosx+b/sqrt(a^2+b^2)*sinx)</p>
<p>a/sqrt(a^2+b^2)
b/sqrt(a^2+b^2)
each of the above numbers have a magnitude less than 1 and their squares add up to 1; therefore one can equal to siny and and the other one to cosy</p>
<p>therefore
sqrt(a^2+b^2)<em>(a/sqrt(a^2+b^2)</em>cosx+b/sqrt(a^2+b^2)<em>sinx)
becomes
sqrt(a^2+b^2)</em>(siny<em>cosx + cosy</em>sinx)
= sqrt(a^2+b^2)*(sin(y+x)</p>
<p>it is clear that it has an aplitude of sqrt(a^2+b^2)</p>
<p>Rusen Meylani</p>
<p>It's one of those Barron's questions (test 3 #50) where you are just supposed to know the answer/formula - there is no way to derive it on test. </p>
<h2>And if you don't know the formula, Barron's serves you well giving a full solution, which you may choose to skip - it's only the result that is VERY important to memorize:</h2>
<h2>a cos x + b sin x = sqrt(a^2 + b^2) * sin(something).</h2>
<p>Amplitude of the sin() is 1, thus
amplitude of the sqrt(a^2 + b^2) * sin () is
sqrt(a^2 + b^2).</p>
<p>Ironically, Meylani gave here a full abstract proof, but in 15 tests for #23 in test 4 his solution is to graph.
Here's the question from 15 tests:
++++++++++++</p>
<h1>What's the amplitude of the f(x) = -3 sin x + 4 cos x + 1.</h1>
<p>It takes you 5 sec or less if you've finished your vegetables, I mean Barron's, first, and if you also know your 3-4-5 triangle:
f(x) = 5 sin (something) + 1.
Amplitude = 5.
+++++++++++++++
In almost unimaginable twist of math 2c, if you get something like
Solve
-3 cos x + 4 sin x = 5,</p>
<h1>go Barron's way:</h1>
<p>-3 cos x + 4 sin x = 5 ( -3/5 cos x + 4/5 sin x) = 5.
There is such t that
sin t = -3/5 and cos t = 4/5 - easy to see on the unit circle.
tan t = -3/4 (so you would not need to worry what quadrant t is in).
t = tan^-1 (-3/4) = -.64
5 ( (sin t) * (cos x) + (cos t) * (sin x) ) = 5
sin(t+x) = 1,
t+x = 1.57
x = 1.57 - t = 1.57 - (-.64) = 2.21.
It'd better be one of the 5 answers!</p>
<p>Just to avoid misunderstanding:
tan^-1 is TI83 key for arctan.</p>
<p>is this on 2c?
yeah, I never saw this kind of problem in my precalc class. we only went over basic trigonometry and sin/cos/tan/csc/sec/cot graphs and stuff like that.</p>
<p>yep, 2c, but i don't anybody who got it for real test.</p>
<p>I don't know why CB would put a question that cannot be easily solved without calculus/memorized formula. I admit, I am having difficulty following the solutions from some of the explanations.</p>
<p>Whats up with bringing back my old threads! I am a guy by the way.</p>
<p>royalties from now on rolling your way stix!
i am just collecting nectar from the old flowers (i am a newbee, hehe).
looking for challenges but not too heavy - not above 2*c speed limit, sir.</p>