Posted: 3/18/2008 5:13:32 PM EDT
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Anyone else not able to make heads or tails of this crap? I have a 5 question quiz due next week and I can't make heads or tails of this shit. |
Idk the first question is "prove (1-2cosx-3cos^2x)/(sin^2x)=(1-3cosx)/(1-cosx)" If I am no mistaken the two denominators equal eachother (sin^2x=1-cosx). |
Took a 22-question test last Friday (this week is spring break) over the same stuff. I'm no genius at it, but passing PreCalc with a high B. Might be able to help. IM me the full question. |
Now simplify the numerator on the left side, 1-2cosx-3cos^2x, until you get 1-3cosx. |
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I have had trouble with trig identities for 4 years now and I'm finally able to deal with them. The only ones you really have to remember are: tan(x) = sin(x)/cos(x) sec(x) = 1/cos(x), csc(x) = 1/sin(x), cot(x) = 1/tan(x) (cos^2)(x) + (sin^2)(x) = 1 From those, you can get pretty much every other one. It can take a little while; it took me about thirty minutes to figure out how to put (tan^4)(x) in terms of sec(x) to solve an integral, though most of that was due to continually making simple mistakes. Edit: for helping with your problem, lemme see if I can guide you through it. First, though, remember that sin^2(x) is NOT 1-cos(x), but 1-cos^2(x). Edit 2: Here's a very good start to that problem: for the left side of the equation, factor the numerator, then put the denominator in terms of cos(x) and factor it. You remember algebra, right? Lemme know if you need help with that, because once you do that it's trivially simple. |
Argh, missed that. I missed almost all the lectures on this stuff due to family medical issues so I have not a clue what the fuck I am doing here. I have no idea how to apply the trig identities. |
How do I go about factoring the top? For the bottom do you mean change sin^2x = (1-cos(2x)/2) ? I have a feeling ur shaking ur head at that...  
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Sin^2(x) = (1-cos^2(x)), remember. To factor, try substituting if you have trouble conceptualizing. Let u = cos(x), so instead of 1 - 2 cos(x) - 3 cos^2(x), it's 1 - 2u - 3u^2, and the denominator is 1 - u^2. Factor those, then replace u with cos(x) again. |
You will need trig identities in Integral Calculus, there are a lot of times where a trigonometric substitution can make integration much easier. But as for more practical stuff like construction, land nav, it doesn't take long. |
when I factor 1 - 2u - 3u^2 I get (u + 1) (u + (1/3))? |
First term is right, but try the second one again. When you multiply it out with the second term you get u^2 + 4/3 u + 1/3. I'd find it easier to do the terms in the factors in the same order they are in the original equation, as in (1 + u) instead of (u + 1). Might help you out a little. Edit: if you want I could just give you a step-by-step, but I figure it'd help you more to solve through yourself. |
A step by step would probably help me more in this instance. I think I am sort of missing the overall concept here. |
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Alright. So, we take your original equation: [(1 - 2 cos(x) - 3 cos^2(x))/sin^2(x)] = [(1 - 3 cos(x))/(1 - cos(x)] Convert the sin^2(x) to (1-cos^2(x)) by the trig identity, and factor the left side of the equation. (1-cos^2(x)) factors to (1 - cos(x))*(1 + cos(x)), and the numerator factors to (1 + cos(x))*(1-3 cos(x)). This yields: [(1 + cos(x))*(1-3 cos(x))/(1-cos(x))*(1 + cos(x))] = [(1 - 3 cos(x))/(1 - cos(x)] You can cancel out the (1+cos(x)) because it exists in both the numerator and the denominator. So you get: [(1-3 cos(x))/(1-cos(x))] = [(1 - 3 cos(x))/(1 - cos(x)] And there you have it. |
