如题!已知cos(π-α)=3/5,α∈[0,π),则sin(2α-π/4)=?要有具体步骤啊,
来源:学生作业帮 编辑:搜狗做题网作业帮 分类:数学作业 时间:2024/06/24 09:19:21
如题!已知cos(π-α)=3/5,α∈[0,π),则sin(2α-π/4)=?要有具体步骤啊,
![如题!已知cos(π-α)=3/5,α∈[0,π),则sin(2α-π/4)=?要有具体步骤啊,](/uploads/image/z/16941485-29-5.jpg?t=%E5%A6%82%E9%A2%98%21%E5%B7%B2%E7%9F%A5cos%28%CF%80-%CE%B1%29%3D3%2F5%2C%CE%B1%E2%88%88%5B0%2C%CF%80%29%2C%E5%88%99sin%282%CE%B1-%CF%80%2F4%29%3D%3F%E8%A6%81%E6%9C%89%E5%85%B7%E4%BD%93%E6%AD%A5%E9%AA%A4%E5%95%8A%2C)
∵cos(π-α)=3/5 ==>-cosα=3/5
∴cosα=-3/5
∵α∈[0,π)
∴sinα=√(1-cos²α)=4/5
故sin(2α-π/4)=sin(2α)cos(π/4)-cos(2α)sin(π/4)
=2sinαcosα*(√2/2)-(2cos²α-1)(√2/2)
=√2(4/5)(-3/5)-√2(-3/5)²+√2/2
=√2(1/2-12/25-9/25)
=-17√2/50
∴cosα=-3/5
∵α∈[0,π)
∴sinα=√(1-cos²α)=4/5
故sin(2α-π/4)=sin(2α)cos(π/4)-cos(2α)sin(π/4)
=2sinαcosα*(√2/2)-(2cos²α-1)(√2/2)
=√2(4/5)(-3/5)-√2(-3/5)²+√2/2
=√2(1/2-12/25-9/25)
=-17√2/50
已知cos(π/6-π)=根号3/3,求cos(5π/6+π)-sin²(α+π/3) 要具体步骤哈- -
已知α∈(0,π/2),且2sinα-sinαcosα-3cosα=0.求[sin(α+π/4)]/[sin2α+cos
已知(3sinα+cosα)/(3cosα-sinα)=2,则2-3sin(α-3π)sin(1.5π-α)-[cos(
已知α∈(0,π/2),且2sin²α-sinαcosα-3cos²α=0,求[sin(α+π/4)
已知sinα+cosα=三分之根号二,sinα-cosα=-4/3,且α∈(-π/2,0),计算(1+sin2α+cos
已知tan(α+π/4)=2,则(2sinα+cosα)/(3cosα-2sinα)
已知6sin²α+5sinαcosα-4cos²α=0,α∈(3π/2,2π),求tanα的值
已知α∈(π,2π) sinα+cosα=1/5 求sinα*cosα sinα-cosα
已知 sinα+2cos(5π/2+α)/cos(π-α)-sin(π/2-α)=-1/4 求(sinα+cosα)平方
已知向量a=(sinα+cosα,√2sinα),b=(cosα-sinα,√2cosα),a∈[0,π/2],且
已知tan(π-α)=2,求sinα-2sinαcosα-cosα/4cosα-3sinα的值
已知sin a=2cos a 计算⑴(sinα+2cosα)/(5cosα-sinα) ⑵tan(α+π/4)