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Finding points of intersection algebraically between 2 trig functions

  1. Jul 15, 2014 #1
    So I have several problems that ask me to find all points of intersection algebraically, but I haven't been able to make much headway on most of them.


    The first problem
    1. The problem statement, all variables and given/known data
    Find all the points of intersection algebraically of the graphs of .... on the interval [0, 4π]

    2. Relevant equations
    f(x) = sinx + 1
    g(x) = cosx

    3. The attempt at a solution

    -Attempt #1:
    sinx + 1 = cosx
    sinx + 1 - cosx = 0
    sinx - cosx = -1

    Deadend

    -Attempt #2:
    sinx + 1 = cosx
    sinx / cosx + 1 / cosx = cosx / cosx
    tanx + secx = 1
    sinx/cosx + 1/cosx = 1

    Deadend

    -Attempt #3:
    sinx + 1 = cosx
    sinx^2 + 1 = cosx^2
    sinx^2 + 1 = 1 - sinx^2
    2sinx^2 + 1 = 1
    2sinx^2 = 0

    ... I think I'm really close to the answer here, but I'm not sure where to go now.


    The second problem
    1. The problem statement, all variables and given/known data
    Find all the points of intersection algebraically of the graphs of .... on the interval [0, 4π]

    2. Relevant equations
    f(x) = tanx
    g(x) = sinx

    3. The attempt at a solution

    sinx = tanx
    sinx = sinx / cosx
    sinx * cosx - sinx = 0
    sinx (cosx - 1) = 0

    sinx = 0
    cosx = 1

    y = 0
    x = 1
    (1,0)

    It's (1,0) at [0, 2π, 4π], but apparently [π, 3π] are also points of intersection, even though the cosine value at [π, 3π] are -1. Could somebody explain that please? (I've pretty much already solved it)


    The third problem
    1. The problem statement, all variables and given/known data
    Find all the points of intersection algebraically of the graphs of .... on the interval [0, 4π]

    2. Relevant equations
    f(x) = tanx
    g(x) = cotx

    3. The attempt at a solution
    Attempt #1:
    tanx = cotx
    sinx / cosx = cosx / sinx
    sinx^2 / cosx = cosx
    sinx^2 = cosx^2
    sinx^2 = -sinx^2 + 1
    2sinx^2 = 1

    And again ... I think I'm really close to the answer here, but I'm not sure where to go now.

    Any helpful advice or tips would be really appreciated! Thankyou.
     
    Last edited: Jul 15, 2014
  2. jcsd
  3. Jul 15, 2014 #2
    Question 1:
    You used the correct method to obtain the result (which is to square both sides and then substitute ##\cos^2x## for ##1-\sin^2x##), however you made a mistake when simplifying since ##(\sin x+1)^2\neq \sin^2 x+1##. To obtain the correct result you may use the following identity: ##(a+b)^2=a^2+2ab+b^2##.

    Question 2:
    ##\sin x=0\,\text{ or }\,\cos x=1## does not mean that ##y = 0, x = 1##, can you see where you made the mistake?

    Question 3:
    Let me help you again: ##2\sin^2x = 1\Rightarrow 2\sin^2x-1=0\Rightarrow\left(\sqrt{2}\sin x-1\right)\left(\sqrt{2}\sin x+1\right)=0.## Do you know how to proceed?
     
    Last edited: Jul 15, 2014
  4. Jul 15, 2014 #3
    Thanks a lot HakimPhilo, you really cleared it up for me, but I still don't understand question 2 though.
    I thought that the only time cosx = 1 was when x = 1 (Likewise for sinx = 0 being y = 1).
     
  5. Jul 15, 2014 #4
    You're welcome. For Q.2 you should understand that we're dealing with units of measure of angles. So when you say ##\cos x=1## means ##x=1## you should specify the unit you used, and since you're surely dealing with radians then for ##x=1## doesn't mean that ##\cos x=1##. (try it on your calculator)
    Instead of ##x=1##, calculate the trigonometric values of an angle with radian measure ##0##.
     
    Last edited: Jul 15, 2014
  6. Jul 15, 2014 #5
    Mmm, I think I'm still a little slow on the uptake. I still don't really understand why [π, 3π] work, because the cos values of [π, 3π] are both -1, and if cosx = 1, then that doesn't make the equation true, since -1 doesn't equal 1. I think I'm really missing some fundamental basic idea here though ....

    cosx = 1
    cos(pi) = -1
    cos(3pi) = -1
     
  7. Jul 15, 2014 #6
    Do you know how to solve equations of the form ##a\cos x+b=0## or ##a\sin x+b=0##? If not then the following may help you a bit:
    ##a\cos x=b\Rightarrow \cos x=\tfrac ba.## What this equation equation means is that we're looking for a number ##x## such that ##\cos x## is equal to ##b/a##. To find this number we can use our classical trigonometric values table:

    http://hotmath.com/hotmath_help/topics/values-of-trigonometric-ratios-for-standard-angles/values-of-trigonometric-ratios-for-standard-angles-image004.gif

    So for example if we want to solve the equation ##2\cos x=\sqrt{2}##, we proceed as follows:
    ##2\cos x=\sqrt{2}\Rightarrow \cos x=\tfrac{\sqrt{2}}2.## We look at our trigonometric table and we find that the angle whose cosine corresponds to ##\tfrac{\sqrt{2}}2## is ##\tfrac\pi4##. So ##\tfrac\pi4## solves our equation. But it's not the only solution, since numbers of the form ##\tfrac\pi4+2n\pi## (where ##n## is an integer) are also solutions.

    If it is not in our table then we use the ##\arccos## function. For example if you have an equation like ##3\cos x=1\Rightarrow \cos x=\tfrac13\,(1)##, you can see that ##\tfrac13## isn't in our table. To solve this problem we use the ##\arccos## function which is like the inverse of ##\cos##, more precisely ##\arccos(\cos x)=x##. So by ##(1)## we have that ##\arccos(\cos x)=\arccos \tfrac13\Rightarrow x=\arccos \tfrac13##. And remember not only ##\arccos \tfrac13## is a solution but also numbers of the form ##\arccos \tfrac13+2n\pi##. (where ##n## is an integer)

    If you need more explanations feel free to ask.

    For the moment I have to sleep...
     
    Last edited: Jul 15, 2014
  8. Jul 15, 2014 #7

    SammyS

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    This is for your problem #2, right?

    You have an equation which you show is equivalent to

    sin(x) (cos(x) - 1) = 0 .

    This equation is satisfied if either of the following conditions is is true.

    sin(x) = 0 OR cos(x) - 1 = 0.

    It doesn't matter that cos(π) = -1 , because sin(π) = 0 .

    What is 0 times -2 ?
     
  9. Jul 15, 2014 #8
    Thank you Sammy for clearing that up! I thought it had to satisfy both equations, since I was finding the point of intersection, but I guess it only needs to satisfy one or the other.

    And thanks again HakImPhilo for the in depth explanation, it really helped clear my understanding.
     
  10. Jul 16, 2014 #9
    Mmm, so just to double check my answers...

    Question 1:
    sinx + 1 = cosx
    (sinx + 1)^2 = 1 - sin^2x
    sin^2x + 2sinx + 1 = 1 - sin^2x
    2sin^2x + 2sinx = 0
    sin^2x + sinx = 0

    x = 0, pi, 2pi, 3pi, 4pi

    (And for the record, how would I input sin^2x into my calculator anyways? I think it's different from sinx^2 right?)

    Question 3:

    ##2\sin^2x = 1\Rightarrow 2\sin^2x-1=0\Rightarrow\left(\sqrt{2}\sin x-1\right)\left(\sqrt{2}\sin x+1\right)=0.##

    sinx = -1 / sqrt(2) → sinx = -sqrt(2) / 2
    sinx = +1 / sqrt(2) → sinx = sqrt(2) / 2

    x = 5pi/4, 7pi/4, 13pi/4, 15pi/4, 3pi/4, pi/4, 11pi/4, 9pi/4

    --------------------------------------------------------
    And on an unrelated note, how would I graph something like f(x) = cos(x+2)? I understand how to graph f(x) = cos(x), but the +2 is really throwing me off (I also understand that it shifts the graph horizontally right by 2, but it feels really awkward graphing it, considering my x-axis goes from 0 - 2pi, in increments of pi/2). Should I just try to approximate it?
     
  11. Jul 16, 2014 #10
    Question 1:
    Try plugging in ##0,\pi,2\pi,3\pi,4\pi## and see if they satisfy your original equation ##\sin x+1=\cos x##.

    You should distinguish between ##\sin^2x## and ##\sin x^2##. The first one is just a notation for ##(\sin x)(\sin x)##, so if you want to calculate it you first find the value of ##\sin x## then you square it. The second one means you calculate the sine of x squared.

    Question 2:
    Yes, all the numbers you listed are solutions to that equation that are in the interval ##[0, 4π]##.

    _________________________________________________________

    Do you know how to graph a function of the form ##f(x)=A\cos(Bx-C)## in general? If not then the following may help you:

    5sKTi.png

    To see the difference between ##f(x)=\cos x ## and ##g(x)=\cos (x+L)##, I've ploted this graphic for various values of ##L##. And as you can see, they shift the graph of ##\cos x## by a horizontal amount of ##L##.


    mcnug.gif
     
    Last edited: Jul 16, 2014
  12. Jul 16, 2014 #11
    Wow, thankyou again HakImPhilo, that diagram really helped!

    Am I still doing problem 1 incorrectly? I seem to get different points of intersection when I graph
    sin^2x + sinx = 0
    compared to sinx + 1 = cosx

    sinx + 1 = cosx
    (sinx + 1)^2 = 1 - sin^2x
    sin^2x + 2sinx + 1 = 1 - sin^2x
    2sin^2x + 2sinx + 1 = 1
    2sin^2x + 2sinx = 0
    sin^2x + sinx = 0
     
  13. Jul 16, 2014 #12
    That's because when you squared both sides of the equations you added extraneous solutions. So you'll have to check the solutions you got and to eliminate those that don't work.
     
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