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Homework Help: Cos√x = cosx

  1. Apr 3, 2012 #1
    1. The problem statement, all variables and given/known data

    cos√x = cosx SOLVE

    2. Relevant equations

    3. The attempt at a solution
    cos√x = cosx
    square both sides
    cos√x*cos√x = cosx*cosx
    cosx = cosx^2
    cosx^2 - cosx = 0

    cosx(cosx - 1) = 0

    cosx = 0 cosx = 1

    x=2Pin, NEI

    x = pi/2 + 2PIn, Nei

    however this is terribly wrong my teacher said, and she told me that the answer is something crazy. and yea.. i'm a newb so please help me.
    Last edited: Apr 3, 2012
  2. jcsd
  3. Apr 4, 2012 #2


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    This line is wrong. What you've done here is square both sides:


    But you've made the left hand side become

    [tex]\left( \cos\sqrt{x}\right) ^2=\cos \left( \sqrt{x}\right) ^2=\cos(x)[/tex]

    Which is wrong.

    [tex]\left( \cos A \right)^2\neq \cos\left(A^2\right)[/tex]

    I'm not exactly sure what answer your teacher is expecting of you (are you supposed to give the general solution which has infinitely many values? Or are you restricted to [itex]0\leq x \leq 2\pi[/itex]?)

    For the general solution, what you should instead be doing is using the fact that if

    [tex]\cos A=\cos B[/tex] then [tex]A=2\pi n \pm B[/tex]

    So for example, if [itex]A=\pi/2[/itex] then [itex]\cos A=0[/itex] but for [itex]\cos B[/itex] to also be zero, B could be [itex]\pi/2, 3\pi/2, 5\pi/2, -\pi/2[/itex], or more generally, [itex]2\pi n \pm \pi/2[/itex] for any integer n.
    Last edited: Apr 4, 2012
  4. Apr 4, 2012 #3
    Of course you could [STRIKE]cheat[/STRIKE] shortcut a little, if you only want one or two solutions... one case in which ##\cos√x = \cos x## is when ##√x = x##.

    But there are many other solutions, which are more difficult to arrive at. To get started on these, you should think about what it means for ##a## and ##b## if ##\cos a = \cos b##.
  5. Apr 4, 2012 #4


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    Never mind. I should read text more carefully. And there doesn't seem to be a way to delete posts in this section.
  6. Apr 4, 2012 #5

    D H

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    That's a nice hint. Here's a huge hint one: Let [itex]u=\surd x[/itex]. With this, the original problem becomes finding the solutions to [itex]\cos(u) = \cos(u^2)[/itex]. Now use Mentallic's hint. You'll get a quadratic equation in u.
  7. Apr 4, 2012 #6
    Alright i think i understand what you guys are saying, and thanks for the great hints.. heres what i've got.

    let√x = u


    cosu = cosu^2

    0 = cosu^2 - cosu
    0= cosu(cosu - 1)

    Cosu = 0 , cosu = 1

    cos√x= 0 cos√x= 1
    √x = ∏/2 √x= 0
    so since

    cosA=cosB, A = 2∏n +/-B

    √x = 2∏n +/- x

    +/-x = √x -2∏n

    x = √x -2∏n or x = -(√x -2∏n)

    x = ∏/2 - 2∏n or x = -(∏/2 - 2∏n)
    Last edited: Apr 4, 2012
  8. Apr 4, 2012 #7


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    Noo... You're breaking the same rule that I mentioned earlier again.

    [tex]\cos(u^2)\neq (\cos(u))^2[/tex]

    So you aren't allowed to factorize [itex]\cos(u^2)[/itex] into [itex]\cos(u)\cdot \cos(u)[/itex]

    Use the rule for [itex]\cos A=\cos B[/itex] that I gave you, and go from there.

    Yes I did :tongue:
    Last edited: Apr 4, 2012
  9. Apr 4, 2012 #8

    D H

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    Plutonium88, you are still making the same mistake you made in your original post. [itex](\cos(u))^2[/itex] and [itex]\cos(u^2)[/itex] are very different things.

    Mentallic beat me to it.
  10. Apr 4, 2012 #9
    Cosx^1/2 = cosx
    Let u = x^1/2

    Cosu = cos(u)^2
    u = 2PIn +\- u^2

    1. -u^2 -u + 2pin =0
    0= u^2 + u - 2pin =0

    2. U^2 - u + 2pin

    1. Quadratic formula

    U = (-1) +\- (1^2 -4(1$(-2pin)^1/2/2

    u=-1 +\- (1^2 -4(1)(-2pin)^1/2/2

    #2 has a negative discriminant.

    1. Simplified to

    U = -1 + (1+8pin)^1/2/2

    Am I on the right track?
  11. Apr 4, 2012 #10


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    Yes you are, so now you just need to substitute [itex]u=\sqrt{x}[/itex] back in and square the expression to find x.

    I personally would have preferred not to use the substitution, and just work from [tex]x=2\pi n\pm \sqrt{x}[/tex] and worked from there. But that's just me :smile:
  12. Apr 4, 2012 #11
    so sub sqrt x back in..

    √x = -1/2 + √(1+8∏n)/2

    square both sides

    x = (-1+ √(1+8∏n))/2^2

    x = (-1 + v(1+8∏n))(-1 + (√1+8∏n))

    x = 8∏n - 2√(1+8∏n) + 2

    Now presuming this is correct, how can i explain this, like explain cosa=cos b where a = 2∏n +/-b

    it looks similiar to the circumfrance of a circle formula 2∏r, but i just dont understand where this relation is from
  13. Apr 4, 2012 #12


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    You forgot about dividing by 4.

    So you want to know why if [itex]\cos a=\cos b[/itex] then [itex]a=2\pi n \pm b[/itex] ?

    If you take a look at the chart that helps determine the signs of the trigonometric function (the cartesian graph that has ASTC in its 4 quadrants respectively) then you know that the cosine function is positive in the 1st and 4th quadrants.

    This means that for an angle [itex]\theta[/itex] above the x-axis in the 1st quadrant, the cosine of that angle is equivalent to the cosine of the same angle [itex]\theta[/itex] made below the x-axis in the 4th quadrant.
    So that means [itex]\cos(\theta)=\cos(-\theta)[/itex] but also we know that the cosine function is periodic with a period of [itex]2\pi[/itex], which means that [tex]\cos(\theta)=\cos(\theta+2\pi)=\cos(\theta-10\pi)[/tex] etc. or more generally,
    [tex]\cos(\theta)=\cos(\theta+2\pi n)[/tex] for any integer n.

    So if we combine both these ideas together, we can come up with the answer to your problem.

    It's futile to try and make sense of the answer in terms of other trigonometric identities that you know of, because the relation isn't simple at all. This is because [itex]\sin\sqrt{x}[/itex] isn't periodic. As x gets large, the wave gets wider and wider (bigger distance between each cycle).
    All the trigonometric conversions you've done in class would have probably involved something along the lines of [tex]A\sin(B+kx)[/tex] for some constants A, B, k. Functions of this type are periodic so their relation to other periodic trig functions can be easily determined.
  14. Apr 4, 2012 #13
    thanks a lot man, all this information has been incredible and i really appreciate you checking my solution, it really helps benefit my math skills.

  15. Apr 5, 2012 #14


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    No problem :smile:
  16. Apr 5, 2012 #15
    x = (8∏n - 2√(1+8∏n) + 2)/4 (divide everything by 2)

    x=( 4∏n - √(1+8∏n) + 1 )/2

    but i showed this to my teacher and she said it's incorrect :'(
    Last edited: Apr 5, 2012
  17. Apr 5, 2012 #16


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    Where's the [itex]\pm[/itex]? :tongue:

    [tex]x=\frac{1+4\pi n\pm \sqrt{1+8\pi n}}{2}[/tex]

    edit: and of course be sure to mention that the only real values of x are for [itex]n\geq 0[/itex]
    Last edited: Apr 5, 2012
  18. Apr 8, 2012 #17
    Much obliged
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