化简sin(2-2π)+cos(2π+α)+tan(α-4π)-cos(10π+α)-sin(α-12π)
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/20 06:56:06
化简sin(2-2π)+cos(2π+α)+tan(α-4π)-cos(10π+α)-sin(α-12π)化简sin(2-2π)+cos(2π+α)+tan(α-4π)-cos(10π+α)-sin(
化简sin(2-2π)+cos(2π+α)+tan(α-4π)-cos(10π+α)-sin(α-12π)
化简sin(2-2π)+cos(2π+α)+tan(α-4π)-cos(10π+α)-sin(α-12π)
化简sin(2-2π)+cos(2π+α)+tan(α-4π)-cos(10π+α)-sin(α-12π)
sin(α-2π)+cos(2π+α)+tan(α-4π)-cos(10π+α)-sin(α-12π)
=sinα+cosα+tanα-cosα-sinα
=tanα
一般地,加减角的2π倍函数名不变,符号不变.
更一般地,如果是π/2的整数倍,可以利用“奇变偶不变,符号看象限”来进行记忆.
即π/2的奇数倍函数名不变,偶数位则函数名改变.
符号要看变化前角所在的象限,如sin(π/2+α)
π/2是π/2的奇数倍,函数名化为cos,再看π/2+α所在的象限(把α看作锐角),则π/2+α在第二象限,第二象限,sin为正,因此sin(π/2+α)=cosα.
sin(α-2π)+cos(2π+α)+tan(α-4π)-cos(10π+α)-sin(α-12π)
=sinα+cosα+tanα-cosα-sinα
=tanα
化简:sin(2π-α)cos(π+α)cos(11π/2-α)/cos(π-α)sin(3π-α)sin(-π-α)sin(9π/2+α)
[sin(5π-α)cos(-π-α)]/[cos(α-π)cos(π/2+α)]化简
化简sin(-π/2-α)sin(πα)cos(-α-π)/cos(π-α)sin(3π α)
(1+sinα+cosα)*[sin(α/2)-cos(α/2)]/√(2+2cosα)化简 (3/2*π
若cosα>sinα,(-π/2
2sinα*cosα*cos(2π-α)+cos(π+2α)*cosα*tanα化简!
化简:sin(-a)cos(2π+a)sin(-a-π)
化简[sin(π/2-α)cos(π/2-α)]/cos(π+α)-[sin(π-α)cos(π/2-α)]/sin(π+α)
化简sin(α-5π)/cos(3π-α)×cos(π/2-α)/sin(α-3π)×cos(8π-α)/sin(-α-4π)
化简sin(α-2π)+sin(-a-3π)cos(α-3π)/cos(π-α)-cos(-π-α)cos(α-4π)
化简cos(3/2π+π)sin(-π+α)
化简:2sin(π+α)cos(π-α)
化简[tan(2π-α)sin(-2π-α)cos(6π-α)]/[cos(α-π)sin(5π-α)]
化简tan(2π-α)sin(-2π-α)cos(6π-α)/cos(α-π)sin(5π-α)
化简sin(π+a)cos(-a)+sin(2π-a)cos(π-a)
化简:[ cos(α-π/2)/sin(5π/2+α) ] ×sin(α-2π) × cos(2π-α)RT.
化简:[1+2sin(α+2π)*cos(α-2π)]/[sin(α+4π)+cos(α+8π)]
化简sin(α-3/2π)cos(α-π)-sin(α-2π)cos(α-π/2)=?