化简[tan(2π-α)sin(-2π-α)cos(6π-α)]/[cos(α-π)sin(5π-α)]
人气:201 ℃ 时间:2020-06-17 20:29:10
解答
[tan(2π-α)sin(-2π-α)cos(6π-α)]/[cos(α-π)sin(5π-α)]=[tan(2π-α)sin(4π-2π-α)cos(4π+2π-α)]/[cos(π-α)sin(4π+π-α)]=[-tanα*sin(2π-α)cos(2π-α)]/[cos(π-α)sin(π-α)]=[-tanα*(-sin...
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