
#1
Apr312, 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. 



#2
Apr412, 12:10 AM

HW Helper
P: 3,436

[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. 



#3
Apr412, 12:35 AM

P: 329

Of course you could
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
Apr412, 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.




#5
Apr412, 05:41 AM

Mentor
P: 14,433





#6
Apr412, 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) 



#7
Apr412, 06:57 AM

HW Helper
P: 3,436

[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 



#8
Apr412, 06:58 AM

Mentor
P: 14,433

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
Apr412, 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? 



#10
Apr412, 09:34 PM

HW Helper
P: 3,436

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 



#11
Apr412, 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 



#12
Apr412, 11:28 PM

HW Helper
P: 3,436

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 1^{st} and 4^{th} quadrants. This means that for an angle [itex]\theta[/itex] above the xaxis in the 1^{st} quadrant, the cosine of that angle is equivalent to the cosine of the same angle [itex]\theta[/itex] made below the xaxis in the 4^{th} 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(\theta10\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. 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
Apr412, 11:40 PM

P: 167

d:D 



#14
Apr512, 01:37 AM

HW Helper
P: 3,436





#15
Apr512, 04:08 PM

P: 167

x=( 4∏n  √(1+8∏n) + 1 )/2 but i showed this to my teacher and she said it's incorrect :'( 



#16
Apr512, 07:44 PM

HW Helper
P: 3,436

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] 



#17
Apr812, 12:45 AM

P: 167

Much obliged



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