<p>Could someone please explain the half-angle identities to me please?</p>
<p>Thanks.</p>
<p>Hey my math teacher helped us memorize this really easy. All you need to know is cos(x/2)</p>
<p>cos(x/2)= sqrt[(1-cosX)/2]</p>
<p>then sin would have the opposite sign. sin(x/2)=sqrt[(1+cosX)/2]</p>
<p>tan(x/2)=sin(X/2)/cos(x/2) = sqrt[(1+cosx)/(1-cosX)] since the 2's cancel</p>
<p>you can get the other two forms by multiplying either the top or bottom by its conjugate sqrt(1+or-cosX). thus 1-cosX^2 = sinX^2. but the sinX^2 is under a radical so you only get sinX.</p>
<p>thus you get: sinX/1-cosX or (1+cosX)/sinX </p>
<p>All have plus or minus depending on where the half of the angle falls and taking the sign or that location. </p>
<p>Also it is good to check on the calculator by findning the angle and playing around with the numbers.</p>
<p>Aiming for 750-800 Saturday. crossing fingers.</p>
<p>Tell me if this was any good help. I really hope so b/c I should be doing hw right now.</p>
<p>Hey!</p>
<p>Thanks so much - that's awesome! :)</p>
<p>Thanks for taking the time to write that up for me!! Appreciate it!</p>
<p>P.S. Good luck on Saturday, Marchballer! I hope you get that 800!</p>