Show that im(f)$ is Möbius strip

Is a parametrization of the Mأ¶bius strip: f:R×(−1/2,1/2)→R3 :(x,y)↦((1+ycos(x/2))cos(x),(1+ycos(x/2))sin(x),ysin(x/2))f:\mathbb R\times\mathbb (-1/2, 1/2)\to\mathbb R^3\ :(x,y)\mapsto ((1+y\cos(x/2))\cos(x),(1+y\cos(x/2))\sin(x),y\sin(x/2))

Show that im(f)$ is a differentiable manifold that is Mأ¶bius strip. I have considered the parameterization as a ruled surface, but not if I have to find the implicit equations Mأ¶bius strip.آ؟Could you tell me how to proceed?.

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Consider using \cos and \sin to get cos\cos and sin\sin instead of cos and sin which give coscos and sinsin.
– Fly by Night
2 days ago

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What exactly do you mean by Moebius strip? Answering this question will help you to understand what should the solution to your problem look like.
– xyzzyz
2 days ago

  

 

Möbius band a ruled surface
– Gabriel Buendia Garcia
2 days ago

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There are lots of ruled surfaces that aren’t Möbius strips. What do you mean by the Möbius strip? How is it defined?
– arkeet
2 days ago

  

 

I define the parameterization f
– Gabriel Buendia Garcia
2 days ago

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