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求值sin10π/3-根号2cos(-19π/4)+tan(13π/3)
人气:269 ℃ 时间:2020-10-01 09:46:50
解答
sin10π/3-根号2cos(-19π/4)+tan(13π/3)
=-sinπ/3-√2cos(3π/4)+tan(π/3)
=-√3/2+1+√3
=1-(√3/2)
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