Thread: Cos√x = cosx
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Apr4-12, 11:28 PM
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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.