已知sin(α+π)=4/5,且sinαcosα<0,求〔2sin(α-π)+3tan(3π-α)〕/4cos(α-3π)的值.
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/11 12:04:07
![已知sin(α+π)=4/5,且sinαcosα<0,求〔2sin(α-π)+3tan(3π-α)〕/4cos(α-3π)的值.](/uploads/image/z/1730039-23-9.jpg?t=%E5%B7%B2%E7%9F%A5sin%28%CE%B1%2B%CF%80%29%3D4%2F5%2C%E4%B8%94sin%CE%B1cos%CE%B1%EF%BC%9C0%2C%E6%B1%82%E3%80%942sin%28%CE%B1-%CF%80%29%2B3tan%283%CF%80-%CE%B1%29%E3%80%95%2F4cos%28%CE%B1-3%CF%80%29%E7%9A%84%E5%80%BC.)
已知sin(α+π)=4/5,且sinαcosα<0,求〔2sin(α-π)+3tan(3π-α)〕/4cos(α-3π)的值.
已知sin(α+π)=4/5,且sinαcosα<0,求〔2sin(α-π)+3tan(3π-α)〕/4cos(α-3π)的值.
已知sin(α+π)=4/5,且sinαcosα<0,求〔2sin(α-π)+3tan(3π-α)〕/4cos(α-3π)的值.
sin(α+π)=4/5,
-sinα=4/5
sinα=-4/5
且sinαcosα<0
cosα=3/5
tanα=sinα/cosα=-4/3
〔2sin(α-π)+3tan(3π-α)〕/4cos(α-3π)
=[-2sinα-3tanα]/(-4cosα)
=(2sinα+3tanα)/(4cosα)
=(-4/5*2-3*4/3)/(4*3/5)
=(-2/5-1)/(3/5)
=(-7/5)*5/3
=-7/3
因为sin(α+π)=-sinα=4/5
所以sinα=-4/5
因为sinαcosα<0
所以cosα=3/5
〔2sin(α-π)+3tan(3π-α)〕/4cos(α-3π)
=〔-2sin(π-a)+3tan(π-α)〕/4cos(π-a)
=(-2sina-3tanα)/(-4cosa)
全部展开
因为sin(α+π)=-sinα=4/5
所以sinα=-4/5
因为sinαcosα<0
所以cosα=3/5
〔2sin(α-π)+3tan(3π-α)〕/4cos(α-3π)
=〔-2sin(π-a)+3tan(π-α)〕/4cos(π-a)
=(-2sina-3tanα)/(-4cosa)
=(2sina+3sinα/cosa)/(4cosa)
=[sina(2cosa+3)]/(4cos^2a)
=[-4/5(2*3/5+3)]/[4*(3/5)^2]
=-7/3
收起