Register to reply

Cos√x = cosx

by Plutonium88
Tags: cos√x, cosx
Share this thread:
Plutonium88
#1
Apr3-12, 10:33 PM
P: 167
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.
Phys.Org News Partner Science news on Phys.org
World's largest solar boat on Greek prehistoric mission
Google searches hold key to future market crashes
Mineral magic? Common mineral capable of making and breaking bonds
Mentallic
#2
Apr4-12, 12:10 AM
HW Helper
P: 3,515
Quote Quote by Plutonium88 View Post
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.
Joffan
#3
Apr4-12, 12:35 AM
P: 361
Of course you could cheat 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##.

K^2
#4
Apr4-12, 02:00 AM
Sci Advisor
P: 2,470
Cos√x = cosx

Never mind. I should read text more carefully. And there doesn't seem to be a way to delete posts in this section.
D H
#5
Apr4-12, 05:41 AM
Mentor
P: 15,065
Quote Quote by Mentallic View Post
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.
Plutonium88
#6
Apr4-12, 06:01 AM
P: 167
Alright i think i understand what you guys are saying, and thanks for the great hints.. heres 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)
Mentallic
#7
Apr4-12, 06:57 AM
HW Helper
P: 3,515
Quote Quote by Plutonium88 View Post
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
D H
#8
Apr4-12, 06:58 AM
Mentor
P: 15,065
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.
Plutonium88
#9
Apr4-12, 06:27 PM
P: 167
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?
Mentallic
#10
Apr4-12, 09:34 PM
HW Helper
P: 3,515
Quote Quote by Plutonium88 View Post
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
Plutonium88
#11
Apr4-12, 10:03 PM
P: 167
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
Mentallic
#12
Apr4-12, 11:28 PM
HW Helper
P: 3,515
Quote Quote by Plutonium88 View Post
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.


Quote Quote by Plutonium88 View Post
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.

Quote Quote by Plutonium88 View Post
it looks similiar to the circumfrance of a circle formula 2∏r, but i just dont 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.
Plutonium88
#13
Apr4-12, 11:40 PM
P: 167
Quote Quote by Mentallic View Post
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
Mentallic
#14
Apr5-12, 01:37 AM
HW Helper
P: 3,515
Quote Quote by Plutonium88 View Post
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
Plutonium88
#15
Apr5-12, 04:08 PM
P: 167
Quote Quote by Mentallic View Post
No problem
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 :'(
Mentallic
#16
Apr5-12, 07:44 PM
HW Helper
P: 3,515
Where's the [itex]\pm[/itex]?

[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]
Plutonium88
#17
Apr8-12, 12:45 AM
P: 167
Much obliged


Register to reply

Related Discussions
Easy exponent precalc problem. 4^(1+√2) * 4^(1-√2) Precalculus Mathematics Homework 4
Analysis Question: Find the supremum and infimum of S,where S is the set S = {√n − [√ Calculus & Beyond Homework 12
Find the supremum and infimum of S, where S is the set S = {√n − [√n]} Calculus & Beyond Homework 8
Prove that (cscx - cotx)^2 = (1-cosx)/(1+cosx) Calculus & Beyond Homework 1
√x-3 + √x = 3 more simple algebra Precalculus Mathematics Homework 11