已知α为第二象限角,sinα+cosα=根号3/3,则cos2α=sinα+cosα=√3/3(sinα+cosα)^2=1/32sinacosa=1/3-1=-2/3sinα-cosα>0,(sinα-cosα)^2=1-2sinacosa=5/3sinα-cosα=√(5/3)为什么cos2α=-(sinα-cosα)(sinα+cosα)=-√(5/3) * (√3/3)=-(
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![已知α为第二象限角,sinα+cosα=根号3/3,则cos2α=sinα+cosα=√3/3(sinα+cosα)^2=1/32sinacosa=1/3-1=-2/3sinα-cosα>0,(sinα-cosα)^2=1-2sinacosa=5/3sinα-cosα=√(5/3)为什么cos2α=-(sinα-cosα)(sinα+cosα)=-√(5/3) * (√3/3)=-(](/uploads/image/z/2533883-59-3.jpg?t=%E5%B7%B2%E7%9F%A5%CE%B1%E4%B8%BA%E7%AC%AC%E4%BA%8C%E8%B1%A1%E9%99%90%E8%A7%92%2Csin%CE%B1%2Bcos%CE%B1%3D%E6%A0%B9%E5%8F%B73%2F3%2C%E5%88%99cos2%CE%B1%3Dsin%CE%B1%2Bcos%CE%B1%3D%E2%88%9A3%2F3%28sin%CE%B1%2Bcos%CE%B1%29%5E2%3D1%2F32sinacosa%3D1%2F3-1%3D-2%2F3sin%CE%B1-cos%CE%B1%3E0%2C%28sin%CE%B1-cos%CE%B1%29%5E2%3D1-2sinacosa%3D5%2F3sin%CE%B1-cos%CE%B1%3D%E2%88%9A%285%2F3%29%E4%B8%BA%E4%BB%80%E4%B9%88cos2%CE%B1%3D-%28sin%CE%B1-cos%CE%B1%29%28sin%CE%B1%2Bcos%CE%B1%29%3D-%E2%88%9A%285%2F3%29+%2A+%28%E2%88%9A3%2F3%29%3D-%28)
已知α为第二象限角,sinα+cosα=根号3/3,则cos2α=sinα+cosα=√3/3(sinα+cosα)^2=1/32sinacosa=1/3-1=-2/3sinα-cosα>0,(sinα-cosα)^2=1-2sinacosa=5/3sinα-cosα=√(5/3)为什么cos2α=-(sinα-cosα)(sinα+cosα)=-√(5/3) * (√3/3)=-(
已知α为第二象限角,sinα+cosα=根号3/3,则cos2α=
sinα+cosα=√3/3
(sinα+cosα)^2=1/3
2sinacosa=1/3-1=-2/3
sinα-cosα>0,
(sinα-cosα)^2=1-2sinacosa=5/3
sinα-cosα=√(5/3)
为什么cos2α=-(sinα-cosα)(sinα+cosα)=-√(5/3) * (√3/3)=-(√5)/3?
已知α为第二象限角,sinα+cosα=根号3/3,则cos2α=sinα+cosα=√3/3(sinα+cosα)^2=1/32sinacosa=1/3-1=-2/3sinα-cosα>0,(sinα-cosα)^2=1-2sinacosa=5/3sinα-cosα=√(5/3)为什么cos2α=-(sinα-cosα)(sinα+cosα)=-√(5/3) * (√3/3)=-(
cos2α
=cos²α-sin²α
=(cosα-sinα)(cosα+sinα)
=-(sinα-cosα)(sinα+cosα)
由公式cos2α=cos^2 α-sin^2 α,因为题中给了sinα+cosα的值,可以求sinα-cosα,于是可以想到用平方差公式cos2α就=-(sinα-cosα)(sinα+cosα)了