Prove that tan n!=0 (n>5)

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Rafe
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Prove that tan n!=0 (n>5)

Unread post by Rafe » Sat Nov 03, 2012 6:43 pm

We know that tan n=0 when n=k.180.6!=1.2.3.4.5.6=180.4.so the factorial of 7,8,9,..... wil be 7.4.180,7.8.4.180.so tan n!=0.(n>5)

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Phlembac Adib Hasan
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Re: Prove that tan n!=0 (n>5)

Unread post by Phlembac Adib Hasan » Sun Nov 04, 2012 10:38 am

Rafe wrote:We know that $\tan n=0$ when $n=k\cdot 180$. $6!=1\times 2\times 3 \times 4 \times 5 \times 6=180\times 4$. So the factorial of $7,8,9,...$ will be $7\times 4\times 180,7\times 8\times 4\times 180$. So $\tan n!=0,(n>5)$
Congrats for an epoch making conclusion. My question is what to do now? Is there anything here for us?
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SANZEED
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Re: Prove that tan n!=0 (n>5)

Unread post by SANZEED » Sun Nov 04, 2012 10:23 pm

Rafe wrote:We know that tan n=0 when n=k.180.6!=1.2.3.4.5.6=180.4.so the factorial of 7,8,9,..... wil be 7.4.180,7.8.4.180.so tan n!=0.(n>5)
Yep,congrats for the result,but please try to use LaTex here. This will make your post look better. See at the announcement section please. :|
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$

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