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Prove that $\left\{ ( \sin a \cos b , \cos a\cos b , \sin b)\mid a , b \in \mathbb{N} \right\}$ is dense in the unit sphere.

Any help will be appreciated.

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What have you tried? Is it the case that when $a$ and $b$ are real numbers, all points of the sphere are covered? – Lubin Sep 3 '12 at 19:14
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Yes,they are basically the spherical coordinates – Roger Sep 3 '12 at 19:23
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@Ricky: consider your last remark: represent any point by spherical coordinates this way, only with $a$ and $b$ real numbers. Can you use density of $\mathbb{Q}$ to first show density of this set with $\mathbb{N}$ replaced with $\mathbb{Q}$? If so, is there a transformation you could use (with perhaps a trigonometric identity) to start with $a,b\in \mathbb{Q}$ and end with $a',b'\in \mathbb{N}$ that gives the same spherical-coordinate point? – Kirk Boyer Sep 3 '12 at 19:32
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Because $\pi$ is irrational, the points $(\cos(n), \sin(n))$ for $n \in \mathbb N$ are dense in the unit circle. And the parametrization $(\sin(a) \cos(b), \cos(a) \cos(b), \sin(b))$ provides a continuous map of the product of two circles onto the sphere. – Robert Israel Sep 3 '12 at 20:05
@Robert Can you just elaborate a bit? – Roger Sep 4 '12 at 5:49
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up vote -4 down vote accepted

Because $\cos(n)$ and $\sin(n)$ (for n an integer) are dense on the interval $[0,1]$ (because $\pi$ is irrational) and those map continu to your coordinates.

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Sorry i did not use tex. Is that the reason for downvote ? – mick Sep 4 '12 at 9:36
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I suspect the reason you got downvotes is that that you basically just copied Robert Israel's comment from above (without even saying so). – Ilmari Karonen Sep 4 '12 at 19:12

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