Tính $\sin {{20}^{0}}\sin {{40}^{0}}\sin {{80}^{0}}$
Tính $\sin {{20}^{0}}\sin {{40}^{0}}\sin {{80}^{0}}$
Started By phathuy, 09-04-2014 - 11:28
#1
Posted 09-04-2014 - 11:28
Mục đích của cuộc sống là sống có mục đích
#2
Posted 09-04-2014 - 11:57
Tính $\sin {{20}^{0}}\sin {{40}^{0}}\sin {{80}^{0}}$
$\sin 20^0.\sin 40^0.\sin 80^0=\frac{1}{2}\left ( \cos 40^0-\cos 120^0 \right )=\frac{1}{2}\left ( \cos 40^0-\frac{1}{2} \right )
=\frac{1}{2}\sin20^0.\cos 40+\frac{1}{4}\sin20^0=\frac{1}{4}(\sin 60^0-\sin 20^0)+\frac{1}{4}\sin 20^0=\frac{1}{4}\sin 60^0=\frac{\sqrt{3}}{8}$
Edited by Kaito Kuroba, 09-04-2014 - 12:06.
- phathuy and megamewtwo like this
#3
Posted 02-08-2015 - 20:33
Ta có: $\sin x\sin(60^0-x)\sin(60+x)=\frac{1}{4}\sin3x$
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