Solving cos√x = cosx: A Newbie's Journey

  • Thread starter Plutonium88
  • Start date
But with \sin\sqrt{x} there is no relation to any other trig functions that you know of, and that's why this problem is so difficult and tedious.But what's important is that you've found the solutions. You've done your job. The explanation is secondary, and I don't think you're expected to understand the explanation, just the solution.
  • #1
Plutonium88
174
0

Homework Statement



cos√x = cosx SOLVE

Homework Equations


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.
 
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  • #2
Plutonium88 said:
cos√x*cos√x = cosx*cosx

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

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

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.
 
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  • #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##.
 
  • #4
Never mind. I should read text more carefully. And there doesn't seem to be a way to delete posts in this section.
 
  • #5
Mentallic said:
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]
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.
 
  • #6
Alright i think i understand what you guys are saying, and thanks for the great hints.. here's what I've got.

let√x = u

cos√x=cosx

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)
 
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  • #7
Plutonium88 said:
0 = cosu^2 - cosu
0= cosu(cosu - 1)

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.

Edit
Yes I did :tongue:
 
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  • #8
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.

Edit
Mentallic beat me to it.
 
  • #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

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?
 
  • #10
Plutonium88 said:
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

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?

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:
 
  • #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 similar to the circumfrance of a circle formula 2∏r, but i just don't understand where this relation is from
 
  • #12
Plutonium88 said:
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
You forgot about dividing by 4.


Plutonium88 said:
Now presuming this is correct, how can i explain this, like explain cosa=cos b where a = 2∏n +/-b
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.

Plutonium88 said:
it looks similar to the circumfrance of a circle formula 2∏r, but i just don't understand where this relation is from
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.
 
  • #13
Mentallic said:
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.

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.

d:D
 
  • #14
Plutonium88 said:
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.

d:D

No problem :smile:
 
  • #15
Mentallic said:
No problem :smile:

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 :'(
 
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  • #16
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]
 
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  • #17
Much obliged
 

1. What is the significance of solving cos√x = cosx?

The equation cos√x = cosx is significant because it helps us understand the behavior of cosine function when the argument is a square root. This type of equation is also known as a trigonometric equation, and solving it can lead to a better understanding of trigonometric identities and their applications in various fields such as physics, engineering, and mathematics.

2. Can this equation be solved algebraically?

Yes, this equation can be solved algebraically by applying trigonometric identities and algebraic manipulations. It involves transforming the equation into a form where we can use the known identities to simplify it and eventually solve for the variable x.

3. Are there any special cases when solving this equation?

Yes, there are special cases when solving this equation. One of the common cases is when the square root is a negative number. In this case, the equation has no real solutions. Another special case is when the argument of cosine is a multiple of π, which can lead to infinitely many solutions.

4. Is there a specific method to solve this equation?

There are various methods to solve this equation, such as using trigonometric identities, graphing, and using a calculator. The best method to use may depend on the given equation and the level of precision required in the solution. It is always recommended to check the solutions obtained using different methods to ensure accuracy.

5. What are the applications of solving cos√x = cosx?

The applications of solving cos√x = cosx include solving problems in physics, engineering, and mathematics involving trigonometric functions. In physics, it can help in analyzing the behavior of waves and oscillations. In engineering, it can be used to design and analyze structures that involve periodic motion. In mathematics, it can be used to simplify complex trigonometric expressions and equations.

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