7@man111: How did u do this sum??
\hspace{-16}$If $\bf{x = \prod_{r=0}^{44}\sin \left(\left(2r+1\right)^0)}$\\\\\\ Then $\bf{x}$ is $\bf{\mathbb{R}}$ational or $\bf{\mathbb{I}}$rrational
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1708\hspace{-16}$Let $\bf{\mathbb{A}=\sin (1^0).\sin (3^0).\sin (5^0)...\sin (87^0).\sin (89^0)}\; $\\\\\\ and $\bf{\mathbb{B}=\sin (2^0).\sin (4^0).\sin (6^0)...\sin (86^0).\sin (88^0)}$\\\\\\ Now $\bf{\mathbb{AB}=\(\sin (1^0).\sin (89^0)\).\(\sin (2^0).\sin(88^0\)...\(\sin (44^0).\sin (46^0)\) .\sin (45^0)}$\\\\\\ $\bf{\mathbb{AB}=\(\sin (1^0).\cos(1^0)\).\(\sin (2^0).\cos (2^0)\)...\(\sin (44^0).\cos (44^0)\) .\sin (45^0)}$\\\\\\ $\bf{\mathbb{AB} = \frac {\sin(2^0)} 2 \frac {\sin(4^0)} 2 ... \frac {\sin(88^0)} 2 \sin(45^0)}$\\\\\\ $\bf{\mathbb{AB}=\frac 1 { 2^{44}}.\mathbb{B}.\sin (45^0)=\frac 1 {2^{44}}.\frac {\sqrt 2} 2 . \mathbb{B}=\frac {\sqrt 2}{2^{45}}.\mathbb{B}} $\\\\\\ Which gives $\boxed{\boxed{\bf{\mathbb{A}=\frac {\sqrt 2}{2^{45}}}}}$
7Nice post @Man111.
similar method like this 1.:
http://www.targetiit.com/iit-jee-forum/posts/trigo-problem-20442.html