解三角方程,sin∧4(β)+cos∧4(β)= 7/8引用sin2β=±1\2 cos2β=±√3/2这就说 明这个2β的终边落在俩条直线上,即为在±π \6±4π\6....没错,sin2β=±1/2,解集2β是±π/6,不过,cos2β=±√3/2,解集2β
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![解三角方程,sin∧4(β)+cos∧4(β)= 7/8引用sin2β=±1\2 cos2β=±√3/2这就说 明这个2β的终边落在俩条直线上,即为在±π \6±4π\6....没错,sin2β=±1/2,解集2β是±π/6,不过,cos2β=±√3/2,解集2β](/uploads/image/z/9303929-17-9.jpg?t=%E8%A7%A3%E4%B8%89%E8%A7%92%E6%96%B9%E7%A8%8B%2Csin%E2%88%A74%EF%BC%88%CE%B2%EF%BC%89%2Bcos%E2%88%A74%EF%BC%88%CE%B2%EF%BC%89%3D+7%2F8%E5%BC%95%E7%94%A8sin2%CE%B2%3D%C2%B11%5C2+cos2%CE%B2%3D%C2%B1%E2%88%9A3%2F2%E8%BF%99%E5%B0%B1%E8%AF%B4+%E6%98%8E%E8%BF%99%E4%B8%AA2%CE%B2%E7%9A%84%E7%BB%88%E8%BE%B9%E8%90%BD%E5%9C%A8%E4%BF%A9%E6%9D%A1%E7%9B%B4%E7%BA%BF%E4%B8%8A%EF%BC%8C%E5%8D%B3%E4%B8%BA%E5%9C%A8%C2%B1%CF%80+%5C6%C2%B14%CF%80%5C6....%E6%B2%A1%E9%94%99%EF%BC%8Csin2%CE%B2%3D%C2%B11%2F2%EF%BC%8C%E8%A7%A3%E9%9B%862%CE%B2%E6%98%AF%C2%B1%CF%80%2F6%EF%BC%8C%E4%B8%8D%E8%BF%87%EF%BC%8Ccos2%CE%B2%3D%C2%B1%E2%88%9A3%2F2%EF%BC%8C%E8%A7%A3%E9%9B%862%CE%B2)
解三角方程,sin∧4(β)+cos∧4(β)= 7/8引用sin2β=±1\2 cos2β=±√3/2这就说 明这个2β的终边落在俩条直线上,即为在±π \6±4π\6....没错,sin2β=±1/2,解集2β是±π/6,不过,cos2β=±√3/2,解集2β
解三角方程,sin∧4(β)+cos∧4(β)= 7/8
引用sin2β=±1\2 cos2β=±√3/2这就说 明这个2β的终边落在俩条直线上,即为在±π \6±4π\6....没错,sin2β=±1/2,解集2β是±π/6,不过,cos2β=±√3/2,解集2β是±π/6吧,cos(π/6)=√3/2,cos(π/3)=1/2
解三角方程,sin∧4(β)+cos∧4(β)= 7/8引用sin2β=±1\2 cos2β=±√3/2这就说 明这个2β的终边落在俩条直线上,即为在±π \6±4π\6....没错,sin2β=±1/2,解集2β是±π/6,不过,cos2β=±√3/2,解集2β
因为sin∧2(β)+cos∧2(β)= 1
所以[sin∧2(β)+cos∧2(β)]∧2=1
sin∧4(β)+cos∧4(β)+2sin∧2(β)*cos∧2(β)=1
===>2sin∧2(β)*cos∧2(β)=1/8
[sin∧2(β)-cos∧2(β)]∧2=sin∧4(β)+cos∧4(β)-2sin∧2(β)*cos∧2(β)=3/4
====>±cos2β=√3/2
===>β= kπ/2±π/12
思路同上,π\6±5π\6 终边,即统一表示为 kπ±π/6 这是2β在除2得β=πk\12+kπ\2 ,
sin2β=±1/2,解集2β是±π/6或±5π\6,不过,cos2β=±√3/2,解集2β是±π/6或±5π\6