<li><p>The graph of r = cos theta intersects the graph of r = sin 2 theta at points?
the answer is (rad(3)/2, pi/6), (0, pi/2), and (-rad(3)/2, 5pi/6)
No idea how to do this one</p></li>
<li><p>An equation in rectangular form equivalent to r^2 = 36 sec 2 theta is
answer) x^2 - y^2 = 36
No idea how to arrive at this one either</p></li>
<li><p>(4rad(3) + 4i) ^ 1/6 =
answer- rad(2)cis(25pi/36)
Couldnt figure out this one either… If anyone could help on any of the above please do so</p></li>
</ol>
<p>where did u get these questions from?</p>
<p>first of all none of these will be on SAT II math IIC </p>
<p>but the first one is solved by going to polar mode on your graphing calc and graphing both quations and seeing where they intersect</p>
<p>the rest I don't have time to explain</p>
<p>barrons book for math sat IIc. Its frickin mad hard</p>
<p>I did the Barron's book, and lost hope for life. Then I got the Princeton Review one, and got a supposed 700. I don't know which one is more realistic, but I have more hope. Barron's is the DEVIL.</p>
<p>lol trsut me, thye would never ask this stuff on SAT II math IIC. its seriously not that hard...the first question of your threee might be like a misc. HARD question on the test...but that's it. The last two, NEVER!!!</p>
<ol>
<li>r = cos(theta), and r = sin(2 theta) = (2) sin(theta)cos(theta)</li>
</ol>
<p>So cos(theta) = (2) sin(theta)cos(theta)
Possibilities:
(a) cos(theta) = 0, which gives theta=pi/2, r = cos(pi/2) = 0</p>
<p>(b) cos(theta) <> 0, so we can simplify the last equation to</p>
<h2> 1 = 2 sin(theta), which is true for theta=(pi/6) or (5 pi/6)</h2>
<ol>
<li>r^2 = 36 sec(2 theta) = 36 / cos( 2 theta)
So r^2 ( cos(2 theta)) = 36
(r^2) (cos(theta)^2 - sin(theta)^2) = 36</li>
</ol>
<p>Since x = r cos(theta) and y = r sin(theta), this is just
x^2 - y^2 = 36</p>
<p>dont worry about it. even if on of these are on the test (which there is a chance that there will be one impossibly hard question), you should just omit it. The 2c curve is friendly enough.</p>
<ol>
<li><p>Use solver.</p></li>
<li><p>Changing from polar to rectangular: x=rcos(theta), y=rsin(theta)</p></li>
<li><p>Use Demoivre's</p></li>
</ol>
<p>what is demoivre's</p>
<p>DeMoivre's theorem is a theorem for complex numbers that states (I wish I had LaTeX):</p>
<p>If z=r(cos(theta) + i sin(theta)), then z^n = r^n(cos(n<em>theta)+i sin(n</em>theta)). </p>
<p>This utilizes the polar representation of complex numbers.</p>
<p>I actually believe nystudent23 has the wrong answer to #3. </p>
<p>If z=4sqrt3 + 4i and we want to find z^(1/6) we first graph z on the complex plane using 4sqrt3 as the real component and 4 as the complex component. Taking |z| we find that |z|=8. Setting up our polar equations we get cos(theta)= sqrt(3)/2 and sin(theta)= 1/2. Therefore, theta= Arg(z)= pi/6.</p>
<p>We have z=8 cis (pi/6), where cis(theta)= cos(theta)+i sin(theta)</p>
<p>Using DeMoivre's, we have z^(1/6) = sqrt(2)*cis(pi\36). That should be right right answer. Someone correct if I made a mistake.</p>
<p>does SAT II test this theorem?</p>
<p>I took math 2c and got an 800 and i dont think this was on it. i don't think it should be on yours either. expect simpler things with complex numbers like find the real part of (2+3i)^2. pretty easy.</p>
<p>are matrices on this test?</p>
<p>biomath i copied this straight out of barrons book. I worked it out and got the same thing, but the book says theres a value known as k that u have to multiply pi/36 with</p>