<p>Do we need to memorize/know this?</p>
<p>What about the remainder theorem, rational root theorem, factor theorem, etc.?</p>
<p>I have never seen the rule asked except in the Barrons book, which is known to contain information you dont really need to know. If a question asks how many roots an equation will have, graph it and see.</p>
<p>i don't know/forget all of those that you mentioned. I got an 800</p>
<p>ok. thanks. barron's was scaring me for a sec</p>
<p>Nah, you don't need to know those. I didn't know any of that stuff and got an 800.</p>
<p>I guess if you knew law of Sins and Cosines, it'd help you. I can imagine a few questions asking for angle etc, where knowing law of Sine and Cosine would be quicker than doing it the way they want you to do it.</p>
<p>Ok. So, is there really anything that we actually need to have memorized(/programmed into our calculators)? Or, like the sat reasoning is everything that we need to know given to us.</p>
<p>obviously it might be easier with other formulas like law of cos but do we NEED to use them for anything?</p>
<p>If you are bringing a graphing calculator, God knows you don't need to know the rational root theorem.</p>
<p>I STRONGLY suggest you go to ticalc.org and look under the math basic programs for 2 or 3 which contain trigonometric identities (the SAT OS has them, too, but they are too slow to access there), then download them to your calculator. Ideally, you have at least the following memorized, or understand the principles behind them and can derive them on the spot.</p>
<p>Sin x = Opposite / Hypotenuse
Cos x = Adjacent / Hypotenuse
Tan x = Opposite / Adjacent
Tan x = (Sin x)/(Cos x)
Csc x = Hypotenuse / Opposite
Csc x = (Sin x)^(-1) = 1/(sin x)
Sec x = Hypotenuse / Adjacent
Sec x = (Cos x)^(-1) = 1/(cos x)
Cot x = Adjacent / Opposite
Cot x = (Cos x)/(Sin x)</p>
<p>1 = (Cos x)^2 + (Sin x)^2
(Cos x)^2 = 1 - (Sin x)^2
(Sin x)^2 = 1 - (Cos x)^2
(Sec x)^2 = 1 + (Tan x)^2
1 = (Sec x)^2 - (Tan x)^2
(Tan x)^2 = (Sec x)^2 - 1
(Csc x)^2 = (Cot x)^2 + 1
1 = (Csc x)^2 - (Cot x)^2
(Cot x)^2 = (Csc x)^2 - 1</p>
<p>Sin (-x) = -Sin x
Cos (-x) = Cos x
Tan (-x) = -tan x
Csc (-x) = -Csc x
Sec (-x) = Sec x
Cot (-x) = -Cot x</p>
<p>Sin ([Pi/2] - x) = Cos x
Cos ([Pi/2] - x) = Sin x
Tan ([Pi/2] - x) = Cot x
Csc ([Pi/2] - x) = Sec x
Sec ([Pi/2] - x) = Csc x
Cot ([Pi/2] - x) = Tan x</p>
<p>(Sin A)/(a) = (Sin B)/(b) = (Sin C)/(c)
(a)/(Sin A) = (b)/(Sin B) = (c)/(Sin C)</p>
<p>(a^2) = (b^2) + (c^2) - (bc)(Cos A)</p>
<p>They never ask the polynomial stuff in great detail. And if they do, a TI-89 does the trick :P</p>
<p>yeah i got an 800 too and i didnt know the rational root theorem or descrate's law of signs. I forgot if I used an 89.</p>
<p>you don't need to know if since you can easily solve the equation with a calculator.</p>