вычислите а)2cos пи/6-ctg пи/3 б)cos 30 sin 60+cos^2 45 в)2 sin 60+ctg(-пи/6). Created by annasologubova. algebra-ru

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annasologubova @annasologubova

July 2022 1 9 Report

вычислите а)2cos пи/6-ctg пи/3 б)cos 30 sin 60+cos^2 45 в)2 sin 60+ctg(-пи/6)

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lilyatomach

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Объяснение:

$2cos\frac{\pi }{6} -ctg \frac{\pi }{3} = 2*\frac{\sqrt{3} }{2} -\frac{\sqrt{3} }{3} =\sqrt{3} -\frac{\sqrt{3} }{3} = \frac{3\sqrt{3} -\sqrt{3} }{3} =\frac{2\sqrt{3} }{3} ;$

б)

$cos30^{0} sin60^{0} +cos^{2} 45^{0} =\frac{\sqrt{3} }{2} *\frac{\sqrt{3} }{2} +(\frac{\sqrt{2} }{2} )^{2} =\frac{3}{4} +\frac{2}{4} =\frac{5}{4} = 1\frac{1}{4} =1,25;$

в)

$2sin 60^{0} +ctg( -\frac{\pi }{6} ) =2*\frac{\sqrt{3} }{2} -\sqrt{3} =\sqrt{3} -\sqrt{3} =0.$

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