两道三角函数化简题.1.a平方cos1140°+b方sin(-1050°)+absin(-810°) 2.分子:sin(-4分之23π)+cos(4分之9π)-tan(3分之7π)+cot(-6分之11π)分母：3cos(-3分之5π)-sin(6分之13π)-4tan(-4分之7π)-5cos5π
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a^2cos1140°+b^2sin(-1050°)+absin(-810°)
=a^2cos(1140°-360°*3)+b^2sin(-1050°+360°*3)+absin(-810°+360°*2)
=a^2cos60°+b^2sin30°+absin(-90°)
=1/2(a^2+b^2-2ab)=1/2(a-b)^2
sin(-23π/4)+cos(9π/4)-tan(7π/3)+cot(-11π/6)
=sin(-23π/4+6π)+cos(9π/4-2π)-tan(7π/3-2π)+cot(-11π/6+2π)
=sin(π/4)+cos(π/4)-tan(π/3)+cot(π/6)
这些都是特殊角 会求吧 查查课本就行了
3cos(-5π/3)-sin(13π/6)-4tan(-7π/4)-5cos5π
=3cos(-5π/3+2π)-sin(13π/6-2π)-4tan(-7π/4+2π)-5cos（5π-4π）
=3cos(π/3)-sin(π/6)-4tan(π/4)-5cosπ
这些也是特殊角