cos11π/4+tan(-7π/6)+sin(-π)+cos(-5π/6)
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cos11π/4+tan(-7π/6)+sin(-π)+cos(-5π/6)cos11π/4+tan(-7π/6)+sin(-π)+cos(-5π/6)cos11π/4+tan(-7π/6)+sin(
cos11π/4+tan(-7π/6)+sin(-π)+cos(-5π/6)
cos11π/4+tan(-7π/6)+sin(-π)+cos(-5π/6)
cos11π/4+tan(-7π/6)+sin(-π)+cos(-5π/6)
cos11π/4+tan(-7π/6)+sin(-π)+cos(-5π/6)
=cos3π/4+tan(-π/6)+sin(π)+cos(5π/6)
=-√2/2-√3/3+0-√3/2
=-√2/2-5√3/6
cos11π/4+tan(-7π/6)+sin(-π)+cos(-5π/6)
sin(-π/6)+cos11π/3+tan(-5π/6)=急
sin4/3π*cos11/6π*tan3/4π=
(sin2π/3+cos11π/6)tan7π/4=
已知点(tan4π/3,cos11π/6) 是θ终边上一点,求tanθ,sinθ,cosθ
tan11π/*6cos(-25π/6)*sin(-23π/6)sin25π/6*cos11π/3+1/4tan(-29π/4)第一题应该是 tan11π/6*cos(-25π/6)*sin(-23π/6)
3(cos11π/6+isin11π/6) 解复数三角形
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(cos11π/6+isin11π/6)^9的代数形式是多少怎样计算?
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比较cos11π/8与sin10π/9的大小,
复数 -根号3+i的三角形式可以表示为A.2(cosπ/6+isinπ/6) B.2(cos11π/6+isinπ11π/6)C.2(cos5π/6+isin5π/6)D.2(cos11π/6-isinπ11π/6)选哪个???
1.比较大小 1)sin(-5π/7)与cos11π/91)sin(-5π/7)与cos11π/92) x=lnπ,y=log以3为底2的对数z=e^-1/2
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tan( x/2+π/4)+tan(x/2-π/4 )=2tanxtan(x/2+π/4)+tan(x/2-π/4)=[tan(x/2)+tan(π/4)]/[1-tan(x/2)tan(π/4)]+[tan(x/2)-tan(π/4)]/[1+tan(x/2)tan(π/4)]=[tan(x/2)+1]/[1-tan(x/2)]+[tan(x/2)-1]/[1+tan(x/2)]=[(tan(x/2)+1)^2-(tan(x/2)-1)^2]/[1-(tan(x
求证(1)1+tanθ/1-tanθ=tan(π/4+θ) (2)1-tanθ/1+tanθ=tan(π/4-θ)
cos9π/4+tan(-7π/6)+sin21π?、