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Homework Help: 3cosx+4sinx=5, confused about R

  1. Feb 15, 2012 #1
    1. The problem statement, all variables and given/known data


    Solve for X.

    2. Relevant equations



    3. The attempt at a solution




    Rsin(θ)= 3








    arctan(3/4)=0.64 rad

    so X=pi/2-0.64+/-2npi

    What confuses me is that I get a different result if I use negative R:


    Are -pi/2 and pi/2 equal?

    Thank You.
  2. jcsd
  3. Feb 15, 2012 #2


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    When you wrote Rsin(x+θ)=5 it was assumed that R>0. A negative R would correspond to a different θ, θ'=θ±pi.

  4. Feb 15, 2012 #3
    Thank You for the response, Ehild. I am just given f(x)=Rsin(x+θ) to solve certain type of trig equation without any explanation as to where it came from. Honestly, when I wrote Rsin(x+θ)=5 I didn't assume that R>0. Just going by formula. So why is is it assumed R>0? And What's that R? Better yet, where did f(x)=Rsin(x+θ) pop out from? Sorry for too many questions.

    I actually asked this question on another forum and someone said R<0 and R>0 work out to the same answer. I'll go get it.
    Last edited: Feb 15, 2012
  5. Feb 15, 2012 #4
    ""pi/2 + M * pi is the same as -pi/2 + N * pi.

    The first one is a set of numbers of the form pi/2 + M * pi, where M runs ofver all integers (i.e. ...-2, -1, 0, 1, 2, ...)...so, these numbers are:

    {...., pi/2 - 3pi, pi/2 - 2pi, pi/2 - pi, pi/2, pi/2 + pi, pi/2 + 2pi, pi/2 + 3pi,...} = {..., -5/2pi, -3/2pi, -pi/2, pi/2, 3/2pi, 5/2pi, 7/2pi,...}

    The second set is comprised of the numbers of the form -pi/2 + N * pi, where N runs over all integers....so, these numbers are:

    {...., -pi/2 - 3pi, -pi/2 - 2pi, -pi/2 - pi, -pi/2, -pi/2 + pi, -pi/2 + 2pi, -pi/2 + 3pi,...} = {..., -7/2pi, -5/2pi, -3/2pi, -pi/2, pi/2, 3/2pi, 5/2pi,...}

    So, you can see that two sets are the same.

    Another way (easier) is to rewrite both sets as

    pi/2 + M * pi =pi (1/2 + M)

    -pi/2 + N * pi = pi (-1/2 + N)

    and, again, notice that 1/2+M defines the same set of numbers as -1/2 + N. ""

    It sure looks intimidating, so I didn't look at it more closely...Will look now.
  6. Feb 15, 2012 #5


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    Allowing both R=5 and R=-5, you would get the same set of solutions.

    Your mistake is that the tangent function is periodic with pi, instead of 2pi.

    So X=pi/2-(0.64±Npi): X=pi/2-0.64, -pi/2-0.64, 3pi/2-0.64, -3pi/2-0.64,...

    If R=-5, X=-pi/2-(0.64±Npi): X=-pi/2-0.64, pi/2-0.64, 3pi/2-0.64, -3pi/2-0.64....

  7. Feb 15, 2012 #6
    Ok, so, is it true that pi/2-(0.64±Npi)=-pi/2-(0.64±Npi) ?
  8. Feb 15, 2012 #7


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    No with the same N. But it is true that for each N there is an M so that pi/2-(0.64+Npi)=-pi/2-(0.64+Mpi) (M and N are both positive and negative integers).

    For example, if N=-2, x= pi/2-0.64+2pi=5/2pi-0.64. 5/2pi-0.64=-pi/2-0.64-Mpi=>M=-3

  9. Feb 15, 2012 #8
    Thanks. Do I need some kind of background to understand this? Something about sets? My textbook doesn't explain R<0 situation, nor does it require me to go over that.
  10. Feb 15, 2012 #9

    3cos(0.930796327)+4sin(0.930796327)= 4.99996936...when R>0

    3cos(-2.21079633)+4sin(-2.21079633)= -4.99996936...when R<0

    Does the solution for the equation when R<0 still hold true because "pi/2 + M * pi is the same as -pi/2 + N * pi" ?

    Also, just because "pi/2 + M * pi is the same as -pi/2 + N * pi" it doesn't mean pi/2=-pi/2 ?

    So does it mean that "pi/2 + M * pi is the same as -pi/2 + N * pi" is something redundant that has nothing to do with solutions of this equation?

  11. Feb 15, 2012 #10


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    I am sorry, I mislead you a bit. There is only a single θ between 0 an 2pi.

    When you have equations of form a cosx + b sinx =c the easiest way to solve them is to write the left hand side in the form Rsin(x+θ). You can do it, so why not.
    The idea comes from the sum law of the sine function: sin(x+θ)=sinx cosθ +cosx sin θ.

    From Rsinθ=3 and Rcosθ=4 it follows that R=5 and sinθ=0.6, cosθ=0.8. There is a single angle between 0 and 2pi for which both are true: θ=0.6435.
    At the same time, Rsin(x+θ)=5 =>sin(x+θ)=1. x+θ=pi/2±2piN, x=-0.6435+pi/2±2piN.

    If you take R=-5, sinθ=-0.6 and cosθ=-0.8, θ=0.6435-pi. -5sin(x+θ)=5, so sin(x+θ)=-1,
    (x+θ)=-pi/2±2piN. x=pi-0.6435-pi/2±2piN=pi/2-0.6435±2piN. The same as for positive R.
    You do not need to consider negative R-s.

    You can apply other methods, for example expressing sinx as ±sqrt(1-cos2x) and solving the equation for cos(x). There will be invalid solutions, you have to replace back the results into the original equation to exclude them.

    Last edited: Feb 15, 2012
  12. Feb 15, 2012 #11
    Aside from the fact that I have no idea how the left side of acosx + bsinx =c gets to be written as Rsin(x+θ), with R>0 (My textbook doesn't show that- just gives the formula Rsin(x+θ) without the provision R>0 to solve the equation of the form acosx + bsinx =c), I am curious as to why R<0 can't be used.
  13. Feb 15, 2012 #12


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    I edited my previous post meanwhile, read it. You can use negative R-s, and you get the same solution. No reason to use both positive and negative R-s.

  14. Feb 16, 2012 #13
    Thank You. I have couple questions:

    1. θ=0.6435-pi. Where did -pi come from?
    2. So, I see how Rsin(x+θ) came to be, except, how did R get there? What is it? What does it do in practice? What does this R really mean? :shy:

    Thank You.
  15. Feb 16, 2012 #14


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    1. Sorry, it is a typo. When sinθ=-0.6 and cosθ=-0.8, θ=0.6435+pi. It comes from the unit circle.

    2.The linear combination of a sine and a cosine of the same angle always equals to a single sine or a cosine of an angle multiplied with a number. In Physics, it appears when two vibrations add up, or two waves interfere. The result is a new vibration or wave, with a new amplitude (it is R) and phase constant (it is theta).
    In Maths, you have two functions f1(x)=acosx, f2(x)=bsinx . If you add them, you get the function g=f1+f2, which has the same shape - a sine or cosine function, but shifted and having a certain amplitude, different from a and b. See attachment.


    Attached Files:

  16. Feb 16, 2012 #15
    sinθ=0.6, θ=0.6435,

    sinθ=-0.6, θ=-0.6435

    How/why does arcsin(-0.6) = -0.6435 become θ=0.6435+pi and arcsin(0.6) is just θ=0.6435, without any pi's attached and/or signs changed?

    Thank You.
  17. Feb 16, 2012 #16
    Awesome explanation. Thanks.

    Is there also an algebraic way to see how R jumps in front of sin(x+θ)=f(x) ?

    Thank You.
  18. Feb 16, 2012 #17


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    R does not jumps:smile: in front of sin(x+θ): you write it there. As the sine function can not exceed 1, you must put a factor R which multiplies sin(x+θ) so as it equals to f(x)=a cosx +b sinx, with arbitrary a and b.

  19. Feb 16, 2012 #18


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    The domain of the arcsin function is [-pi/2, pi/2], that of the cosine function is [0, pi].
    The sin and cosine do not determine the angle completely. sin(θ)=sin(pi-θ), so the equation sinθ=-0.6 has two groups of solutions, θ=arcsin(-0.6)+2Npi =-0.6435+2Npi and pi-arcsin(-0.6)+2Npi=pi+0.6435+2Npi. Between 0 and 2pi you have 2pi-0.6435=5.6397 and 3.7851.

    cos(θ)=cos(-θ). Therefore the equation cos(θ)=-0.8 has two solutions: θ=arccos(-0.8)+2Npi and θ=-arccos(-0.8)+2Npi. arccos(-0.8)=2.4981. Between 0 and 2pi the solutions are 2.4981 and 2pi-2.4981=3.7851.

    So θ=3.7851 is the angle which cosine is -0.8 and sine is -0.6.

    In case sinθ=0.6 and cosθ=0.8, arcsin(0.6)=0.6435, so θ=0.6435 + 2Kpi or θ=pi-0.6435 + 2Kpi=2.4981+2Kpi.
    arccos(0.8)=0.6435, so θ=0.6435 + 2Kpi or θ=-0.6435 + 2Kpi, that is 5.6397 between 0 and 2pi. The common solution is θ=0.6435 + 2Kpi.

    Draw the unit circle and see the relation between angles which have the same cosine or sine.

  20. Feb 16, 2012 #19
    Thank You, Ehild.
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