已知α∈(π/2,π),且sinα=3/5 则(1)cos(α-π/4) (2)求sin平方α/2+tan(α+π/4)
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已知α∈(π/2,π),且sinα=3/5 则(1)cos(α-π/4) (2)求sin平方α/2+tan(α+π/4)的值
已知α∈(π/2,π),且sinα=3/5
则(1)cos(α-π/4) (2)求sin平方α/2+tan(α+π/4)的值
已知α∈(π/2,π),且sinα=3/5
则(1)cos(α-π/4) (2)求sin平方α/2+tan(α+π/4)的值
![已知α∈(π/2,π),且sinα=3/5 则(1)cos(α-π/4) (2)求sin平方α/2+tan(α+π/4)](/uploads/image/z/5875332-60-2.jpg?t=%E5%B7%B2%E7%9F%A5%CE%B1%E2%88%88%28%CF%80%2F2%2C%CF%80%29%2C%E4%B8%94sin%CE%B1%3D3%2F5+%E5%88%99%281%29cos%28%CE%B1-%CF%80%2F4%29+%282%29%E6%B1%82sin%E5%B9%B3%E6%96%B9%CE%B1%2F2%2Btan%EF%BC%88%CE%B1%2B%CF%80%2F4%EF%BC%89)
α∈(π/2,π),且sinα=3/5
cosα=-4/5
cos(α-π/4)
=cosαcosπ/4+sinαsinπ/4
=-4/5*√2/2+3/5*√2/2
=-4√2/10+3√2/10
=-√2/10
α∈(π/2,π)
α/2∈(π/4,π/2)
cosα=-4/5
cosα=1-2sin²α/2
-4/5=1-2sin²α/2
-9/5=-2sin²α/2
sin²α/2=9/10
tanα=-3/4
(2)sin²α/2+tan(α+π/4)
=9/10+(tanα+tanπ/4)/(1-tanαtanπ/4)
=9/10+(-3/4+1)/[1-(-3/4)]
=9/10+(1/4)/[7/4]
=9/10+1/7
=63/70+10/70
=73/70
再问: 谢谢。
再答: sin2α/2 =(1-cosα)/2
cosα=-4/5
cos(α-π/4)
=cosαcosπ/4+sinαsinπ/4
=-4/5*√2/2+3/5*√2/2
=-4√2/10+3√2/10
=-√2/10
α∈(π/2,π)
α/2∈(π/4,π/2)
cosα=-4/5
cosα=1-2sin²α/2
-4/5=1-2sin²α/2
-9/5=-2sin²α/2
sin²α/2=9/10
tanα=-3/4
(2)sin²α/2+tan(α+π/4)
=9/10+(tanα+tanπ/4)/(1-tanαtanπ/4)
=9/10+(-3/4+1)/[1-(-3/4)]
=9/10+(1/4)/[7/4]
=9/10+1/7
=63/70+10/70
=73/70
再问: 谢谢。
再答: sin2α/2 =(1-cosα)/2
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