13-14----和1求解,求过程
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13.解:原式=cos²28°+cos²(90°-28°)+ tan33°·tan(90°-33°)
=cos²28°+sin²28°+ tan33°·cot33°
=1+1
=2
14.解:原式
=tan(-180°+30°)cos(-180°-30°)cos(360°+60°)/[cos(-720°+120°)sin(-360°+30°)]
=tan30°cos30°cos60°/[cos120°sin30°]
=sin30°cos60°/[cos60°sin30°]
=1
1.解:tan(11π/3)=tan(4π - π/3 )
=tan( π/3 )
=√3
=cos²28°+sin²28°+ tan33°·cot33°
=1+1
=2
14.解:原式
=tan(-180°+30°)cos(-180°-30°)cos(360°+60°)/[cos(-720°+120°)sin(-360°+30°)]
=tan30°cos30°cos60°/[cos120°sin30°]
=sin30°cos60°/[cos60°sin30°]
=1
1.解:tan(11π/3)=tan(4π - π/3 )
=tan( π/3 )
=√3
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