一道数学题,题目如图所示。要求写出详细解答过程!我先谢谢大神们了!在线等!
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(1)sinx在[π/2,π]递减
所以sin2π/3﹥sin4π/5
(2)sin(-110°)=-sin(70°)
sin(-820°)=sin(-100°)=-sin(80°)
∵sin(70°)<sin(80°)
∴-sin(70°)>-sin(80°)
∴sin(-110°)>sin(-820°)
(3)sin(314°)=sin(360°-46°)=-sin(46°)<0
sin(769°)=sin(720°+49°)=sin(49°)>0
所以 (314°)〈sin(769°)
所以sin2π/3﹥sin4π/5
(2)sin(-110°)=-sin(70°)
sin(-820°)=sin(-100°)=-sin(80°)
∵sin(70°)<sin(80°)
∴-sin(70°)>-sin(80°)
∴sin(-110°)>sin(-820°)
(3)sin(314°)=sin(360°-46°)=-sin(46°)<0
sin(769°)=sin(720°+49°)=sin(49°)>0
所以 (314°)〈sin(769°)
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(1)sin2π/3=sin120°
sin4π/5=sin144
由于sinx在[90°,180°]上单调递减,所以sin120°>sin144°,也就是sin2π/3>sin4π/5
(2)sin(-820°)=sin(-820°+360°×2)=sin(-100°)
由于sinx在[-180°,-90°]上单调递减,所以sin(-110°)>sin(-100°),也就是sin(-110°)>sin(-820°)
(3)sin(-314°)=sin(-314°+360°)=sin(46°)
sin(-769°)=sin(769°-360°×2)=sin(49°)
由于sinx在[0°,90°]上单调递减,所以sin49°>sin46°,也就是sin(-314°)< sin(-769°)
(4)sin(-54π/7)=sin(-54π/7+28×2π/7)=sin(2π/7)
sin(-63π/8)=sin(-63π/8+32×2π/8)=sin(π/8)
由于2π/7∈[0,π/2],π/8∈[0,π/2],2π/7>π/8且sinx在[0,π/2]单调递增,所以sin(2π/7)>sin(π/8),那么也就是sin(-54π/7)>sin(-63π/8)
望采纳哦!!!
sin4π/5=sin144
由于sinx在[90°,180°]上单调递减,所以sin120°>sin144°,也就是sin2π/3>sin4π/5
(2)sin(-820°)=sin(-820°+360°×2)=sin(-100°)
由于sinx在[-180°,-90°]上单调递减,所以sin(-110°)>sin(-100°),也就是sin(-110°)>sin(-820°)
(3)sin(-314°)=sin(-314°+360°)=sin(46°)
sin(-769°)=sin(769°-360°×2)=sin(49°)
由于sinx在[0°,90°]上单调递减,所以sin49°>sin46°,也就是sin(-314°)< sin(-769°)
(4)sin(-54π/7)=sin(-54π/7+28×2π/7)=sin(2π/7)
sin(-63π/8)=sin(-63π/8+32×2π/8)=sin(π/8)
由于2π/7∈[0,π/2],π/8∈[0,π/2],2π/7>π/8且sinx在[0,π/2]单调递增,所以sin(2π/7)>sin(π/8),那么也就是sin(-54π/7)>sin(-63π/8)
望采纳哦!!!
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