Hi! :) is there a way to get the answer in radical form of can anyone help me?

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Hi! :) is there a way to get the answer in radical form of can anyone help me?

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Hi! :) is there a way to get the answer in radical form of $\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\frac{{\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{2}}}}}{{\mathtt{2}}}}\right)} = {\mathtt{0.041\: \!111\: \!761\: \!829}}$ can anyone help me?

Guest Apr 27, 2015

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sin(3pi/2/2) ?

$\sin_{(rad)} \left(\dfrac{ \dfrac{3\cdot \pi}{2} }{2} \right) =\sin_{(rad)} \left( \dfrac{3\cdot \pi}{4} \right)=\frac{1}{2}\sqrt{2}\\\\ \dfrac{3\cdot \pi}{4}\; \rm{rad} \equiv 135 \ensurement{^{\circ}}\\\\ \sin_{(360 \ensurement{^{\circ}} )} \left(135 \ensurement{^{\circ}} \right) = \frac{1}{2}\sqrt{2}\\\\$

$\alpha$ (°)	$\alpha$ (rad)	$\sin \alpha$	$\cos \alpha$	$\tan \alpha$	$\cot \alpha$
$0^\circ$	$\,0$	$\,0$	$\,1$	$\,0$	$\pm\infty$
$15^\circ$	$\tfrac{\pi}{12}$	$\tfrac14(\sqrt{6}-\sqrt{2})$	$\tfrac14(\sqrt{6}+\sqrt{2})$	$2-\sqrt{3}$	$2+\sqrt{3}$
$18^\circ$	$\tfrac{\pi}{10}$	$\tfrac{1}{4}\left(\sqrt{5}-1\right)$	$\tfrac{1}{4}\sqrt{10+2\sqrt{5}}$	$\tfrac{1}{5}\sqrt{25-10\sqrt{5}}$	$\sqrt{5+ 2\sqrt{5}}$
$30^\circ$	$\tfrac{\pi}{6}$	$\tfrac12$	$\tfrac12\sqrt3$	$\tfrac13\sqrt3$	$\sqrt3$
$36^\circ$	$\tfrac{\pi}{5}$	$\tfrac{1}{4} \sqrt{10- 2\sqrt{5}}$	$\tfrac{1}{4} \left (1+ \sqrt{5} \right)$	$\sqrt{5- 2\sqrt{5}}$	$\tfrac{1}{5} \sqrt{25+ 10\sqrt{5}}$
$45^\circ$	$\tfrac{\pi}{4}$	$\tfrac12\sqrt2$	$\tfrac12\sqrt2$	$1\,$	$1\,$
$54^\circ$	$\tfrac{3\pi}{10}$	$\tfrac{1}{4} \left (1+ \sqrt{5} \right)$	$\tfrac{1}{4} \sqrt{10- 2\sqrt{5}}$	$\tfrac{1}{5} \sqrt{25+ 10\sqrt{5}}$	$\sqrt{5- 2\sqrt{5}}$
$60^\circ$	$\tfrac{\pi}{3}$	$\tfrac12\sqrt3$	$\tfrac12$	$\sqrt3$	$\tfrac13\sqrt3$
$72^\circ$	$\tfrac{2\pi}{5}$	$\tfrac{1}{4} \sqrt{10+ 2\sqrt{5}}$	$\tfrac{1}{4} \left (\sqrt{5}-1 \right)$	$\sqrt{5+ 2\sqrt{5}}$	$\tfrac{1}{5} \sqrt{25 - 10\sqrt{5}}$
$75^\circ$	$\tfrac{5\pi}{12}$	$\tfrac14(\sqrt{6}+\sqrt{2})$	$\tfrac14(\sqrt{6}-\sqrt{2})$	$2+\sqrt{3}$	$2-\sqrt{3}$
$90^\circ$	$\tfrac{\pi}{2}$	$\,1$	$\,0$	$\pm\infty$	$\,0$
$108^\circ$	$\tfrac{3\pi}{5}$	$\tfrac{1}{4} \sqrt{10+ 2\sqrt{5}}$	$\tfrac{1}{4} \left (1- \sqrt{5} \right)$	$-\sqrt{5+ 2\sqrt{5}}$	$-\tfrac{1}{5} \sqrt{25 - 10\sqrt{5}}$
$120^\circ$	$\tfrac{2\pi}{3}$	$\tfrac12\sqrt3$	$-\tfrac12$	$-\sqrt3$	$-\tfrac13\sqrt3$
$135^\circ$	$\tfrac{3\pi}{4}$	$\tfrac12\sqrt2$	$-\tfrac12\sqrt2$	$-1\,$	$-1\,$
$180^\circ$	$\pi\,$	$\,0$	$\,-1$	$\,0$	$\pm\infty$
$270^\circ$	$\tfrac{3\pi}{2}$	$\,-1$	$\,0$	$\pm\infty$	$\,0$
$360^\circ$	$2\pi$	$\,0$	$\,1$	$\,0$	$\pm\infty$

heureka Apr 27, 2015

3 Online Users

heureka · Answer 1 · 2015-04-27T06:18:14

sin(3pi/2/2) ?

$\sin_{(rad)} \left(\dfrac{ \dfrac{3\cdot \pi}{2} }{2} \right) =\sin_{(rad)} \left( \dfrac{3\cdot \pi}{4} \right)=\frac{1}{2}\sqrt{2}\\\\ \dfrac{3\cdot \pi}{4}\; \rm{rad} \equiv 135 \ensurement{^{\circ}}\\\\ \sin_{(360 \ensurement{^{\circ}} )} \left(135 \ensurement{^{\circ}} \right) = \frac{1}{2}\sqrt{2}\\\\$

$\alpha$ (°)	$\alpha$ (rad)	$\sin \alpha$	$\cos \alpha$	$\tan \alpha$	$\cot \alpha$
$0^\circ$	$\,0$	$\,0$	$\,1$	$\,0$	$\pm\infty$
$15^\circ$	$\tfrac{\pi}{12}$	$\tfrac14(\sqrt{6}-\sqrt{2})$	$\tfrac14(\sqrt{6}+\sqrt{2})$	$2-\sqrt{3}$	$2+\sqrt{3}$
$18^\circ$	$\tfrac{\pi}{10}$	$\tfrac{1}{4}\left(\sqrt{5}-1\right)$	$\tfrac{1}{4}\sqrt{10+2\sqrt{5}}$	$\tfrac{1}{5}\sqrt{25-10\sqrt{5}}$	$\sqrt{5+ 2\sqrt{5}}$
$30^\circ$	$\tfrac{\pi}{6}$	$\tfrac12$	$\tfrac12\sqrt3$	$\tfrac13\sqrt3$	$\sqrt3$
$36^\circ$	$\tfrac{\pi}{5}$	$\tfrac{1}{4} \sqrt{10- 2\sqrt{5}}$	$\tfrac{1}{4} \left (1+ \sqrt{5} \right)$	$\sqrt{5- 2\sqrt{5}}$	$\tfrac{1}{5} \sqrt{25+ 10\sqrt{5}}$
$45^\circ$	$\tfrac{\pi}{4}$	$\tfrac12\sqrt2$	$\tfrac12\sqrt2$	$1\,$	$1\,$
$54^\circ$	$\tfrac{3\pi}{10}$	$\tfrac{1}{4} \left (1+ \sqrt{5} \right)$	$\tfrac{1}{4} \sqrt{10- 2\sqrt{5}}$	$\tfrac{1}{5} \sqrt{25+ 10\sqrt{5}}$	$\sqrt{5- 2\sqrt{5}}$
$60^\circ$	$\tfrac{\pi}{3}$	$\tfrac12\sqrt3$	$\tfrac12$	$\sqrt3$	$\tfrac13\sqrt3$
$72^\circ$	$\tfrac{2\pi}{5}$	$\tfrac{1}{4} \sqrt{10+ 2\sqrt{5}}$	$\tfrac{1}{4} \left (\sqrt{5}-1 \right)$	$\sqrt{5+ 2\sqrt{5}}$	$\tfrac{1}{5} \sqrt{25 - 10\sqrt{5}}$
$75^\circ$	$\tfrac{5\pi}{12}$	$\tfrac14(\sqrt{6}+\sqrt{2})$	$\tfrac14(\sqrt{6}-\sqrt{2})$	$2+\sqrt{3}$	$2-\sqrt{3}$
$90^\circ$	$\tfrac{\pi}{2}$	$\,1$	$\,0$	$\pm\infty$	$\,0$
$108^\circ$	$\tfrac{3\pi}{5}$	$\tfrac{1}{4} \sqrt{10+ 2\sqrt{5}}$	$\tfrac{1}{4} \left (1- \sqrt{5} \right)$	$-\sqrt{5+ 2\sqrt{5}}$	$-\tfrac{1}{5} \sqrt{25 - 10\sqrt{5}}$
$120^\circ$	$\tfrac{2\pi}{3}$	$\tfrac12\sqrt3$	$-\tfrac12$	$-\sqrt3$	$-\tfrac13\sqrt3$
$135^\circ$	$\tfrac{3\pi}{4}$	$\tfrac12\sqrt2$	$-\tfrac12\sqrt2$	$-1\,$	$-1\,$
$180^\circ$	$\pi\,$	$\,0$	$\,-1$	$\,0$	$\pm\infty$
$270^\circ$	$\tfrac{3\pi}{2}$	$\,-1$	$\,0$	$\pm\infty$	$\,0$
$360^\circ$	$2\pi$	$\,0$	$\,1$	$\,0$	$\pm\infty$

Hi! :) is there a way to get the answer in radical form of can anyone help me?

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Hi! :) is there a way to get the answer in radical form of can anyone help me?

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