f(α)=[sin(π-α)cos(α-π)tan(-α+π)]/[tan(π+α)cos(3/2π+α)] 化简

来源:学生作业帮助网 编辑:作业帮 时间:2024/08/06 20:18:51
f(α)=[sin(π-α)cos(α-π)tan(-α+π)]/[tan(π+α)cos(3/2π+α)] 化简
x)K8Q683O|._ =ߠY2cA PFNӞi5$S0;6)@6jb@@Z_L4! l ޱshe `kCy`Xށh

f(α)=[sin(π-α)cos(α-π)tan(-α+π)]/[tan(π+α)cos(3/2π+α)] 化简
f(α)=[sin(π-α)cos(α-π)tan(-α+π)]/[tan(π+α)cos(3/2π+α)] 化简

f(α)=[sin(π-α)cos(α-π)tan(-α+π)]/[tan(π+α)cos(3/2π+α)] 化简
f(α)=[sin(π-α)cos(α-π)tan(-α+π)]/[tan(π+α)cos(3/2π+α)]
=sinα(-cosα)(-tanα)/(tanαsinα)
=cosα

f(α)=[sin(π-α)cos(α-π)tan(-α+π)]/[tan(π+α)cos(3/2π+α)]
=[sinα cosα tanα]/[tanα sinα]
=cosα