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Learning Trig on my own... I'm stuck and need assitance! 
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#1
Aug2104, 09:09 PM

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Hi everyone. My main goal this week is to learn derivatives of trig functions. But before that I'm still learning some of the trig identities and solving some equations. I'm stuck on a couple of questions that I need help on.
1. Express [tex] \sec 2 \Theta [/tex] in terms of [tex] \sec \Theta [/tex] and [tex] \tan \Theta [/tex] 2. Express [tex] \csc 2 \Theta [/tex] in terms of [tex] \csc \Theta [/tex] and [tex] \sec \Theta [/tex] 3. Express [tex] \sin 2 \Theta[/tex] and [tex] \cos 2 \Theta [/tex] in terms of [tex] \tan \Theta [/tex] 4. Determine [tex] \sin 2 \Theta [/tex] if [tex] \sin \Theta + \cos \Theta = \frac{1}{2}[/tex] 5. Are there any sample problems on the web about finding an expression of a cone within a sphere and all you are given is the radius of the sphere and the semivertical angle of the cone? I have the same question in my book but I cant figure it out. I could write it up here but I cant include the diagram it shows with it. Thanks for any help. 


#2
Aug2104, 11:58 PM

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1. Are you familiar with the double angle formulas?
[tex] \sin2\theta = 2\sin\theta \cos\theta [/tex] Rewrite this expression as: [tex] 2\left[\frac{\sin\theta}{\cos\theta} \cos\theta \right]\cos\theta [/tex] Now you can replace [tex] \inline{\frac{\sin\theta}{\cos\theta}} [/tex] with [tex] \inline{\tan\theta} [/tex] and every [tex] \inline{\cos\theta} [/tex] with [tex] \inline{\frac{1}{\sec\theta}} [/tex] I think that gives you [tex] \frac{2\tan\theta}{\sec^2\theta} [/tex] Now you have an idea how to do these problems. That should get you started. For instance, 2, 3, and 4 all seem to make use of the same identity, though I only gave them a cursory glance. Hint for 4: Square the expression: [tex] (\sin\theta + \cos\theta)^2 = \frac{1}{4} [/tex] When you expand, you're set! 


#3
Aug2204, 12:12 AM

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Yes I have learned the double angle formulas for sin, cos and tan.
The answer provided for number 1. in my answer sheet (just answer, they give no solution) is [tex] \sec 2 \Theta = \frac{\sec^2 \Theta}{1  \tan^2 \Theta} [/tex] I understand what you are trying to show me, but what I am confused on is why do you do it that way? I mean, out of all the things you can do to manipulate the formula, why do you go about it that particular way? Is it the only way? I am trying to understand how do you envision the final answer when the question simply states write in terms of sec and tan. How do I see it? 


#4
Aug2204, 12:35 AM

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Learning Trig on my own... I'm stuck and need assitance!
To answer your second question, I sympathize. I've often wondered..."how the hell did they arrive at that method for solving the problem." It wasn't even remotely obvious. It seemed to come out of nowhere. In this instance however, I attacked the problem as follows: We're starting out with: [tex] \sin2\theta = 2\sin\theta \cos\theta [/tex] We have an expression in terms of sines and cosines, and we want to express it in terms of tangents and secants. Doesn't it make sense to make use of the relationship between the latter pair and the former? How does tangent relate to sine and cosine? How does secant? As you well know: [tex] \tan\theta = \frac{\sin\theta}{\cos\theta} [/tex] [tex] \sec\theta = \frac{1}{\cos\theta} [/tex] Once you know that, you have the first step to replacing your sines and cosines with secants and tangents. Remember that there are usually many ways of approaching a problem, all valid, perhaps some more efficient than others. 


#5
Aug2204, 12:42 AM

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Egads! Sorry! I got confused. 1. is [tex] \sec 2 \Theta [/tex] in terms of [tex]\sec \Theta and \tan \Theta , not \sin 2 \Theta [/tex]
Typing out that special formating for math symbols is rather slow and then my mind wanders where i'm not aware of which trig function im' inputting. Sorry again. But thanks for what you have given me so far. 


#6
Aug2204, 12:45 AM

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I guess it means the solution I gave isn't much help to you for that particular question though. I hope that you've got some insight into the methods now...don't hesistate to ask if you run into more trouble.



#7
Aug2204, 12:48 AM

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I still don't know though what the double angle formulas for sec, csc and cot are, let alone being able to rewrite them in terms of other trig. Are they derivable from the sin, cos, tan ones? And I dont know if there are addition/subtraction formulas for them either.



#8
Aug2204, 06:46 PM

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First of all, just to clear this up: all of the double angle formulas can be derived from the addition/subtraction formulas. They are not really distinct, but just special cases:
e.g.: Addition Formula: [tex] \sin(\theta + \phi) = \sin\theta\cos\phi + \cos\theta\sin\phi [/tex] In the special case where [tex] \inline{ \theta = \phi} [/tex], you get the double angle formula for sine: [tex] \sin(\phi + \phi) = \sin(2\phi) = \sin\phi\cos\phi + \cos\phi\sin\phi = 2\sin\phi\cos\phi [/tex] Now...on to your question of whether such formulas exist for cosecant, secant, and cotangent. I have to ask...why don't you just try deriving them and see! e.g. Secant double angle formula derviation (I just tried this on paper) [tex] \sec2\theta = \frac{1}{\cos2\theta} [/tex] Now, the double angle formula for cosine has three forms. We can write: [tex] \frac{1}{\cos2\theta} = \frac{1}{\cos^2\theta \sin^2\theta} = \frac{1}{2\cos^2\theta  1} = \frac{1}{12\sin^2\theta} [/tex] (If you're not sure how to arrive at the last two forms from the first one, simply take the identity [tex] \inline {\sin^2\theta + \cos^2\theta = 1} [/tex], solve for one or the other, and substitute.) Now, which of the three should we choose? I chose the third one because it has [tex] \inline {\sin^2\theta} [/tex] in it, which is useful. It looks to me that we could turn that expression into one with tangents and secants simply by dividing through by [tex] \inline {\cos^2\theta} [/tex]: [tex] \frac{1}{12\sin^2\theta} = \frac{\frac{1}{\cos^2\theta}}{\frac{1}{\cos^2\theta}  2\tan^2\theta} = \frac{\sec^2\theta}{\sec^2\theta  2\tan^2\theta} [/tex] Now employ the identiy [tex] \inline{ \sec^2\theta = 1 + \tan^2\theta} [/tex]. Make sure you know where it comes from too. With it, that last expression becomes: [tex] \frac{\sec^2\theta}{1+ \tan^2\theta  2\tan^2\theta} = \frac{\sec^2\theta}{1 \tan^2\theta } [/tex] Voila! We derived a double angle formula for secant from the double angle formula for cosine. It follows that you should now be able to derive double angle formulas for cosecant and cotangent from those for sine and tangent respectively. Have fun! 


#9
Aug2304, 11:13 AM

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cepheid:
It is perhaps more "elegant" to use: [tex]\sec(2\theta)=\frac{\sec^{2}(\theta)}{2\sec^{2}(\theta)}[/tex] since it only uses the secant function. 


#10
Aug2304, 11:32 AM

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Cool! Thanks for pointing that out. I suppose that as a double angle identity for secant, that is a better choice. But I was also considering the question as originally posed to mathemagician in his book:



#11
Aug2304, 11:47 AM

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Yes, you're right.
Your answer is the one they're after..(however, my answer also expresses sec(2w) in terms of sec(w) and tan(w) (or [tex]tan^{0}(w)[/tex] anyways )) 


#12
Aug2504, 12:02 AM

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Thanks for the help. I have been able to finish those questions. I have now moved on to limits of Trig functions. Does anyone know of a tutoral on it? I don't think the examples in my my text are adequate. I'm stuck on a couple of limits that I don't understand.
1. [tex] \lim_{\substack{x\rightarrow 0}} \frac{\sin^2 3x }{4x} [/tex] The answer provided is 0. But I can't see how the reach it. Plugging in 0 directly will give you 0/0 and I have been taught that when you get that you have to use a different method to solve the limit. 2. [tex] \lim_{\substack{x\rightarrow 0}} \frac{\sin^2 3x }{4x^3} [/tex] Answer is does not exist. I do not understand. And could someone verify if what I did is right for 3. 3. [tex] \lim_{\substack{x\rightarrow 0}} \frac{\sin^2 3x }{4x^2} [/tex] Solution : [tex] \lim_{\substack{x\rightarrow 0}} \frac{\sin^2 3x }{4x^2} [/tex] = [tex] \lim_{\substack{x\rightarrow 0}} \frac{\ (sin3x)(sin3x) }{(2x)(2x)} [/tex] = [tex] \lim_{\substack{x\rightarrow 0}} \frac{\ (sin3x)(sin3x) }{(2x)(2x)} ( \frac{(\frac{3}{2})^2}{(\frac{3}{2})^2}) [/tex] = [tex] \frac{9}{4}\lim_{\substack{\frac{9}{4}x\rightarrow 0}} \frac{\ (sin3x)(sin3x) }{(3x)(3x)} [/tex] Since sin3x/3x = 1... = [tex] \frac{9}{4}(1)(1) = \frac{9}{4} [/tex] which is the answer they give me. Thanks 


#13
Aug2504, 12:11 AM

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For #1 and #2, you have to use the same method as you did in #3: you have to pair each "sin kx" with a "kx".



#14
Sep604, 12:08 AM

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Thank you. I have finally figured them out.
I am now stuck on another question. [tex]\lim_{x\rightarrow 0} \frac{\cos^2x  1}{x^2}[/tex] Solution 1 (which I believe is correct): [tex]\lim_{x\rightarrow 0} \frac{\sin^2x}{x^2}[/tex] = [tex]\lim_{x\rightarrow 0} \frac{(\sin x)(\sin x)}{(x)(x)}[/tex] = [tex] (1)(1) = 1 [/tex] Solution 2 (Why would this be incorrect?): [tex]\lim_{x\rightarrow 0} \frac{(\cos x  1)(\cos x + 1)}{(x)(x)} = 0[/tex] Thanks 


#15
Sep604, 03:59 AM

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Write [tex]\cos{x}\approx{1\frac{x^{2}}{2}}[/tex] Inserting this, we evaluate the limit as: [tex]\lim_{x\to0}\frac{((1\frac{x^{2}}{2})1)(2\frac{x^{2}}{2})}{x^{2}}=\lim_{x\to0}1+\frac{x^{4}}{4}=1[/tex] 


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