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Trigonomentry, differentiation + equation.

  1. Oct 29, 2013 #1
    I have another two problems I find difficult. They both involve trigonometry, so I thought I could fit both under the same post. Also, if possible, I'd like some help in regards to confirming that one problem I've solved is done correctly.

    1. The problem statement, all variables and given/known data

    First, the derivative. Find y' when y = ##arctan^2(x)##

    2nd problem. Solve the equation ##sin^2x##+[itex]\frac{\sqrt{3}}{2}[/itex]sin2x = 1- ##2cos^2x##, xe[0, 2[itex]\pi[/itex])

    Lastly, I've solved this problem, and hopefully someone can take a quick look at it and verify if I've done it correctly: find the limit when x->∞ (y - [itex]\sqrt{y^2-y}[/itex])
    2. Relevant equations

    I'm not sure what to write here. Relevant equations are how trigonometric functions relate to each other. For example ##sin^2x + cos^2x## = 1. There are a lot of them, and I think it would be confusing to write them all.

    3. The attempt at a solution

    First problem:
    Find y' when y = ##arctan^2(x)##

    y = arctanx * arctanx
    y' = ([itex]\frac{1}{x^2+1}[/itex])##^2##

    I feel that this is way too easy, although it makes sense in a way.

    Edit: in the above problem, does arctanx*arctanx translate to ##arctanx^2## and not ##arctan^2x## as I've written? That might explain a lot, and if that's true, please let me know how I can begin to attempt to solve this problem.

    Problem 2:
    Solve the equation ##sin^2x##+[itex]\frac{\sqrt{3}}{2}[/itex]sin2x = 1- ##2cos^2x##, xe[0, 2[itex]\pi[/itex])

    ##sin^2x##+##2cos^2x## + [itex]\frac{\sqrt{3}}{2}[/itex]sin2x=1

    ##sin^2x##+##2cos^2x## + [itex]\frac{\sqrt{3}}{2}[/itex](2sinxcosx)=1

    ##sin^2x##+##cos^2x##+##cos^2x##+[itex]\frac{\sqrt{3}}{2}[/itex]2(sinxcosx)=1

    ##cos^2x##+[itex]\frac{\sqrt{3}}{2}[/itex]2(sinxcosx)=0

    cosx+[itex]\frac{\sqrt{3}}{2}[/itex]2sinx=0

    cosx+[itex]\sqrt{3}[/itex]sinx=0

    And here I am stuck. I'm not even sure what the answer is going to be. I'm imagining I should be getting something like x= [itex]\pi[/itex] + [itex]\pi[/itex]n -> ne[+,2[itex]\pi[/itex]) or something. I'm really confused about this, unfortunately.

    3rd 'problem':
    x->∞ (y - [itex]\sqrt{y^2-y}[/itex])

    [itex]\frac{(y-\sqrt{y^2-y})(y+\sqrt{y^2-y})}{y+\sqrt{y^2-y}}[/itex]

    [itex]\frac{y}{y+\sqrt{y^2-y}}[/itex] -> divide by y.

    [itex]\frac{1}{1+\frac{\sqrt{y^2-y}}{y}}[/itex]

    [itex]\frac{1}{1+\sqrt{\frac{y^2-y}{y^2}}}[/itex]

    using l'hopitals

    [itex]\frac{1}{1+\sqrt{\frac{2y-1}{2y}}}[/itex]

    = [itex]\frac{1}{2}[/itex]

    Any help is appreciated, and just let me know if I can do something different with how I post or something. I'm new here so I still have a lot to learn in regards to the forum as well as math.
     
  2. jcsd
  3. Oct 29, 2013 #2

    LCKurtz

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    Yes, it seems way too easy because it is incorrect. The derivative of a product is not the product of the derivatives. You have something (##\arctan##) raised to a power. You need to start with the power rule followed with the chain rule (derivative of the inside).

    Writing ##arctan x^2## is ambiguous. Much better to write either ##\arctan^2(x)## or ##\arctan(x^2)##, depending on which you mean. Of course the product of two arctangents would be written the first way.

    Can you express this in terms of the tangent?

    Your third problem looks OK.
     
  4. Oct 29, 2013 #3
    Alright. Sorry for not being absolutely clear in my opening post. ##\arctan^2(x)## is the correct problem.

    This was what I was afraid of. I don't know how to start solving this problem. ##\arctan^2(x)## is problematic for me, because I see the problem, black and white. I don't know how to visualize the problem, and I think that makes it harder. In addition to that, I can't find anything in my book that tells me how to rewrite this expression. Only tanx, and even then, I end up with ##\sin^2(x)## and that gets me nowhere I'm afraid.

    Any guidelines on this would be highly appreciated.


    The short answer is no. The longer version is that I don't understand what you mean. I've tried to solve the problem as far as I can with my knowledge, but I struggle with trigonometry because I rarely see the connection with real numbers, so I often think 'should the answer be in numbers, or in trigonometric expressions?'.

    I'm sorry that I don't understand more, but I will work a lot with this tomorrow.

    Thanks for your input!
     
  5. Oct 29, 2013 #4

    Mark44

    Staff: Mentor

    You can find the derivative of [arctan(x)]2 using either the product rule or the chain rule. Why is arctan2(x) (= [arctan(x)]2) problematic for you? If you have a graphing calculator you can view the graph of y = arctan2(x) .

    Regarding LCKurtz's question, it was with regard to this equation:
    cos(x) +√3 * sin(x) = 0.

    To write this in terms of tan(x), divide both sides by cos(x).
     
  6. Oct 30, 2013 #5
    I'm not sure why it was difficult, but I understand it now - I think. Graphing it made it better, thanks.

    This is my thinking:

    (fg)' = f'g + fg'

    f= arctanx f'= [itex]\frac{1}{1+x^2}[/itex]

    g= arctanx g'= [itex]\frac{1}{1+x^2}[/itex]

    = [itex]\frac{1}{1+x^2}[/itex] * arctanx + arctanx * [itex]\frac{1}{1+x^2}[/itex]

    = [itex]\frac{arctanx}{1+x^2}[/itex] + [itex]\frac{arctanx}{1+x^2}[/itex]

    = [itex]\frac{2arctanx}{1+x^2}[/itex]

    First problem solved? At least it makes sense to me this time.

    In regards to the 2nd problem, I did this:

    [itex]\frac{cosx}{cosx}[/itex] +[itex]\sqrt{3}[/itex] [itex]\frac{sinx}{cosx}[/itex] = 0

    1+ [itex]\sqrt{3}[/itex] tanx = 0

    tanx = [itex]\frac{1}{\sqrt{3}}[/itex]

    Now I've attempted to find the solution, and I feel that I'm very close, but I can't remember how we used to do this (I really should repeat exercises more often). I've found out that I should take the arctanx on both sides, but only when the equation is one-to-one function. Unfortunately, I don't understand what this means.

    Am I on to something here?

    Thanks a lot for your help, I really appreciate it. I feel I'm learning a lot.
     
  7. Oct 30, 2013 #6

    Mark44

    Staff: Mentor

    First problem: yes, that's correct.
    Second problem: you're almost done. When you're learning right triangle trig, there are a very small number of angles whose sine, cosine, tangent, etc. you need to memorize. This is one of them.
     
  8. Oct 30, 2013 #7
    Ah! I didn't realize [itex]\frac{1}{\sqrt{3}}[/itex] = [itex]\frac{\sqrt{3}}{3}[/itex] straight away.

    So tanx= 30° -> tanx= [itex]\frac{\pi}{6}[/itex]. To find all solutions within this function, I take tanx + [itex]\pi[/itex] * n until I have all solutions within xe[0,2[itex]\pi[/itex])?

    In that case: x = [itex]\frac{\pi}{6}[/itex] and [itex]\frac{\pi}{6}[/itex] + [itex]\pi[/itex] * 1 = [itex]\frac{7}{6}[/itex][itex]\pi[/itex].

    Once again, I really appreciate your help.
     
  9. Oct 30, 2013 #8

    Mark44

    Staff: Mentor

    I didn't notice earlier, but for the 2nd problem your solution is incorrect. Starting from 1 + √3 tan(x) = 0, you should get tan(x) = - 1/√3.

    Also, some of what you wrote doesn't make sense, putting aside for the moment that you have the wrong value.
    It's incorrect to say tanx = 30° or tanx = ##\pi/6##. You can say tan(30°) = tan(##\pi/6##) = 1/√3.

    It's also incorrect to say tanx = n##\pi##. Instead you should say x = n##\pi##. Do you see the difference?
     
    Last edited: Oct 30, 2013
  10. Oct 30, 2013 #9
    Yes, I can see that now. I made a careless mistake when I subtracted by 1 on both sides of the equation. I've fixed that in my notes now, thanks for noticing!

    Hmm. I guess I've worked too little with these kind of problems, or it's been too long since I last worked on these kind of problems!

    So, if I write tan(-30°) = tan(-[itex]\frac{\pi}{6}[/itex]) = - [itex]\frac{\sqrt{3}}{3}[/itex] that's correct?

    And to then solve my problem in regards to xe[0, 2[itex]\pi[/itex]), I use the equation x = tanx + n[itex]\pi[/itex]?

    Since tanx = - [itex]\frac{\pi}{6}[/itex] I get the following:

    x = - [itex]\frac{\pi}{6}[/itex] + [itex]\pi[/itex] and x = - [itex]\frac{\pi}{6}[/itex] + 2[itex]\pi[/itex] -> x = [itex]\frac{5}{6}[/itex] [itex]\pi[/itex] and x = [itex]\frac{11}{6}[/itex] [itex]\pi[/itex]

    I must admit, I'm still confused about this, but now it makes slightly more sense.

    Thanks a lot for helping me once again, appreciated.
     
  11. Oct 30, 2013 #10

    Mark44

    Staff: Mentor

    Yes.
    No. You are solving the equation tan(x) = -√3/3 to find the value of x. What you're doing is writing the values of x in the interval [0, ##2\pi##] for which tan(x) = -√3/3.
    No. tan(x) = -√3/3, so x = ##-\pi/6##. However, there are an infinite number of x values for which tan(x) = -√3/3. You're concerned only with the values that are in the interval [0, ##2\pi##].

    Your work below looks fine.
     
  12. Oct 30, 2013 #11
    I honestly don't understand this, I'm sorry. Maybe I should take a break and look at it with a fresh mind.


    Oh, I've been mixing up degrees/radians and the value of x. Even if I see that, I'm still confused.

    I will take a break, look at some other exercises (maybe I'll have to post them on here as well - in a new thread of course) and I'll get back to this tonight. I'm not sure why I can't understand this. Hopefully a break will do me good.

    Again, thank you so much for all your help.
     
  13. Oct 30, 2013 #12

    Mark44

    Staff: Mentor

    Because that's not what the problem was. You've been mixing up the input values with the function or output values.

    For the tan function:
    Input value - an angle, which could be in degrees or in radians.
    Output value/function value - a number.

    Some examples:

    Input: 30°
    Output: 1/√3
    This says that tan(30°) = 1/√3

    Input: ##\pi/6##
    Output: 1/√3
    This says that tan(##\pi/6##) = 1/√3
    Note that 30° and ##\pi/6## represent exactly the same angle.

    Input: 45°
    Output: 1
    This says that tan(45°) = 1

    Input: 60°
    Output: √3
    This says that tan(60°) = √3

     
  14. Oct 31, 2013 #13
    For your second problem there is one easy way. There a basic rule for differentiating functions such as f(g(x)) known as the chain rule. The chain rule is basically differentiating the functions and multiplying them. Well, in case of f(g(x)) the differentiation will be f'(g(x)).g'(x)

    For eg. sin^2(x) (Similar to f(g(x)) )

    In the above equation f(x) is a^2 where a=sinx and g(x) is sinx.
    ∴ The differetiation will be 2.sin(x).cos(x)


    Now for your question, which is arctan^2(x).

    Here, f(x) is a^2 and g(x) is arctan x.

    Hence the differentiation will be 2arctanx.1/1+x^2
     
  15. Oct 31, 2013 #14
    See, x here is basically the angle, which is mostly taken in radian. What I mean is,##arctanx^2## and ##arctan^2x## are two completely different things. When you multiply two trigonometric functions your answer cannot affect the angle. I hope I cleared this doubt?
     
  16. Oct 31, 2013 #15

    Mark44

    Staff: Mentor

    Parentheses would help. arctan(x) * arctan(x) is the same as (arctan(x))2. Another way to write this would be arctan2(x), understanding that it means (arctan(x))2.

    arctanx2 is pretty ambiguous, as it's not clear whether x is being squared or arctan(x) is being squared. A nonambiguous version would be arctan(x2).
     
  17. Oct 31, 2013 #16

    LCKurtz

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    Is it just me, or is this thread going around in circles?
     
  18. Oct 31, 2013 #17

    Mark44

    Staff: Mentor

    It's not you - for the benefits of Mutaja and phyneach I restated pretty much what you said back in post #2.
     
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