不查表求值:cos5π/12·sinπ/12=1.如题2.根号下(cos4-sin^2(2)+2)=3.若sinα=4/5,tan(α+β)=1,α为第二象限角,则tanβ=4.sin2x/2cosx(1+tanx·tan(x/2))化简结果为
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![不查表求值:cos5π/12·sinπ/12=1.如题2.根号下(cos4-sin^2(2)+2)=3.若sinα=4/5,tan(α+β)=1,α为第二象限角,则tanβ=4.sin2x/2cosx(1+tanx·tan(x/2))化简结果为](/uploads/image/z/5412230-62-0.jpg?t=%E4%B8%8D%E6%9F%A5%E8%A1%A8%E6%B1%82%E5%80%BC%EF%BC%9Acos5%CF%80%2F12%C2%B7sin%CF%80%2F12%3D1.%E5%A6%82%E9%A2%982.%E6%A0%B9%E5%8F%B7%E4%B8%8B%28cos4-sin%5E2%282%29%2B2%29%3D3.%E8%8B%A5sin%CE%B1%3D4%2F5%2Ctan%28%CE%B1%2B%CE%B2%29%3D1%2C%CE%B1%E4%B8%BA%E7%AC%AC%E4%BA%8C%E8%B1%A1%E9%99%90%E8%A7%92%2C%E5%88%99tan%CE%B2%3D4.sin2x%2F2cosx%281%2Btanx%C2%B7tan%28x%2F2%29%29%E5%8C%96%E7%AE%80%E7%BB%93%E6%9E%9C%E4%B8%BA)
不查表求值:cos5π/12·sinπ/12=1.如题2.根号下(cos4-sin^2(2)+2)=3.若sinα=4/5,tan(α+β)=1,α为第二象限角,则tanβ=4.sin2x/2cosx(1+tanx·tan(x/2))化简结果为
不查表求值:cos5π/12·sinπ/12=
1.如题
2.根号下(cos4-sin^2(2)+2)=
3.若sinα=4/5,tan(α+β)=1,α为第二象限角,则tanβ=
4.sin2x/2cosx(1+tanx·tan(x/2))化简结果为
不查表求值:cos5π/12·sinπ/12=1.如题2.根号下(cos4-sin^2(2)+2)=3.若sinα=4/5,tan(α+β)=1,α为第二象限角,则tanβ=4.sin2x/2cosx(1+tanx·tan(x/2))化简结果为
1 ,cos5π/12·sinπ/12
=cos(π/4+π/6)*sin(π/4-π/6)
=(√6-√2)/4*(√6+√2)/4=1/4
2.√[cos4-sin^2(2)+2]
=√[cos^2(2)-2sin^2(2)+2]
=√[cos^2(2)+2(1-sin^2(2)]
=√[3cos^2(2)]=-√3cos2
3;sinα=4/5,cosα=-3/5 ,tanα=-4/3
tan(α+β)=1=(tanα+tanβ)/(1-tanαtanβ)
(1-tanαtanβ)=(tanα+tanβ)
1+4/3tanβ=-4/3tanβ ,tanβ=-3/8
4
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不查表求值:cos5π/12·sinπ/12=1.如题2.根号下(cos4-sin^2(2)+2)=3.若sinα=4/5,tan(α+β)=1,α为第二象限角,则tanβ=4.sin2x/2cosx(1+tanx·tan(x/2))化简结果为
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