<p>Any tips/tricks? I mean the double/half angle stuff and the sin/cos/tan(A+B) identities. Do we even need to know them by heart to secure a 800?</p>
<p>I got an 800, but it was very easy for me because it was the weekend before my math final so all the formulas were fresh in my mind already. I don’t remember needing to use a lot of the addition/subtraction/double-angle/half-angle ones though. Make sure you’re solid with the basic identities (Pythagorean, etc.)</p>
<p>Which formulas did you opt to memorize? Also, is it legal to have the formulas saved in the calculator? lol</p>
<p>This link says the double/half-angle formulas are optional though, but I don’t want to risk omitting an easy question just because I was too lazy to try to memorize the needed formula.
<a href=“http://www.erikthered.com/tutor/facts-and-formulas-2b.pdf[/url]”>http://www.erikthered.com/tutor/facts-and-formulas-2b.pdf</a></p>
<p>Any other input?</p>
<p>Another way to do it is to memorize the addition formulas for sin(A+B), cos(A+B), tan(A+B), which shouldn’t be too difficult. Then the double-angle formulas easily follow, as you can let A=B (so if you somehow forget on a test, you can easily derive them). I never remember the half-angle formulas, but some of them can be derived from other formulas, and it’s unlikely you’ll need to use them on Math II.</p>
<p>You should definitely know the identity sin^2(A) + cos^2(A) = 1 (this <em>is</em> the Pythagorean theorem), as well as similar identities such as tan^2(A) + 1 = sec^2(A).</p>
<p>^thanks for the tip! If you don’t mind, how do you derive the half-angle formulas?</p>
<p>Here’s an example for sin(A/2):</p>
<p>We can use the double-angle identity cos A = cos^2(A/2) - sin^2(A/2) = 1 - 2 sin^2(A/2). </p>
<p>With a little algebra this rearranges to sin^2(A/2) = (1 - cos A)/2, so sin (A/2) = ± sqrt{(1 - cos A)/2} (the sign depends on where A/2 is in the unit circle).</p>
<p>@MIT: how would you solve this problem: <a href=“http://talk.collegeconfidential.com/sat-subject-tests-preparation/1480167-google-page-ranking-math-sat-ii-question.html[/url]”>http://talk.collegeconfidential.com/sat-subject-tests-preparation/1480167-google-page-ranking-math-sat-ii-question.html</a></p>
<p>Move everything to one side:
sin 2A + cos 2A - tan A = 0</p>
<p>This is equivalent to
sin 2A cos A + cos 2A cos A - sin A = 0</p>
<p>2 sin A cos^2 A + (1 - 2 sin^2 A) cos A - sin A = 0</p>
<p>Rearranging a bit,</p>
<p>2 sin A cos A (cos A - sin A) + (cos A - sin A) = 0</p>
<p>(cos A - sin A)(sin 2A + 1) = 0</p>
<p>Either cos A - sin A = 0 (answer choice B) or sin 2A = -1. If sin 2A = -1, then A = 3pi/4. This satisfies the original constraint, but if A = 3pi/4, sin A - cos A = sqrt(2). Therefore the answer is E) It cannot be determined from the information given.</p>
<p>I got an 800 on Math Level 2 and I didn’t have to use any of those double and half angle identities. Memorizing those formulas is not incredibly difficult. Just look up proofs for each of the identities.</p>
<p>Proving the addition formulas (sin (A+B), etc.) is somewhat tricky, but not too difficult. However many of the other identities can easily be derived from these.</p>
<p>Lol I’m screwed. It took me like 15 minutes to come up with a solution that matches yours. But:</p>
<p>cos A - sin A = 0</p>
<p>Therefore
sin A - cos A = 0</p>
<p>Because this is all the question asks for, isn’t B the answer?</p>
<p>Don’t worry, that problem wasn’t too obvious for me either. </p>
<p>Not quite! From (cos A - sin A)(sin 2A + 1) = 0 (or (sin A - cos A)(sin 2A + 1) = 0), this implies that <em>either</em> sin A - cos A = 0 or sin 2A + 1 = 0. Unless the solution to the second eqn. satisfies the first eqn., you’re going to get two different values for sin A - cos A.</p>
<p>To check, notice that A = pi/4 and A = 3pi/4 are solutions to the original equation. But sin A - cos A = 0 and sqrt(2) respectively.</p>
<p>You can store anything in your calculator(including the formulas). I did but I didn’t use a single one of them either. Just type them up and you’ll be fine.</p>
<p>Yes, and not often on a test will you have to know more obscure trig identities.</p>