求值:(1)sin(4π/3)cos(7π/6)tan(21π/4)(2)sin(23π/6)cos(-17π/4)tan(π/8)tan(3π/8)要求详细过程
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(1)sin(4π/3)cos(7π/6)tan(21π/4)
=sin(π+π/3)cos(π+π/6)tan(5π+π/4)
=-sinπ/3*(-cosπ/6)tanπ/4
=-√3/2*(-√3/2)*1
=3/4
(2)sin(23π/6)cos(-17π/4)tan(π/8)tan(3π/8)
=sin(4π-π/6)cos(-4π-π/4)tan(π/8)tan(3π/8)
=sinπ/6cosπ/4tan(π/8)tan(3π/8)
=1/2*√2/2*1
=√2/4
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=sin(π+π/3)cos(π+π/6)tan(5π+π/4)
=-sinπ/3*(-cosπ/6)tanπ/4
=-√3/2*(-√3/2)*1
=3/4
(2)sin(23π/6)cos(-17π/4)tan(π/8)tan(3π/8)
=sin(4π-π/6)cos(-4π-π/4)tan(π/8)tan(3π/8)
=sinπ/6cosπ/4tan(π/8)tan(3π/8)
=1/2*√2/2*1
=√2/4
如果您认可我的回答，请点击“采纳为满意答案”,谢谢！
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