怎样把下列三角函数转化为y=Asin(ωx+φ)形式(1)y=sin[2X+(π/8)]cos2X 结果为:y=1/2sin〔4X+(π/8)]+1/2sin(π/8)(2)y=cos^2 * 4X 结果为:y=(1+cos8X)/2
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/01 23:55:27
![怎样把下列三角函数转化为y=Asin(ωx+φ)形式(1)y=sin[2X+(π/8)]cos2X 结果为:y=1/2sin〔4X+(π/8)]+1/2sin(π/8)(2)y=cos^2 * 4X 结果为:y=(1+cos8X)/2](/uploads/image/z/3859143-15-3.jpg?t=%E6%80%8E%E6%A0%B7%E6%8A%8A%E4%B8%8B%E5%88%97%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0%E8%BD%AC%E5%8C%96%E4%B8%BAy%3DAsin%28%CF%89x%2B%CF%86%29%E5%BD%A2%E5%BC%8F%281%29y%3Dsin%5B2X%2B%28%CF%80%2F8%29%5Dcos2X+%E7%BB%93%E6%9E%9C%E4%B8%BA%3Ay%3D1%2F2sin%E3%80%944X%2B%28%CF%80%2F8%29%5D%2B1%2F2sin%28%CF%80%2F8%29%282%29y%3Dcos%5E2+%2A+4X+%E7%BB%93%E6%9E%9C%E4%B8%BA%EF%BC%9Ay%3D%281%2Bcos8X%29%2F2)
怎样把下列三角函数转化为y=Asin(ωx+φ)形式(1)y=sin[2X+(π/8)]cos2X 结果为:y=1/2sin〔4X+(π/8)]+1/2sin(π/8)(2)y=cos^2 * 4X 结果为:y=(1+cos8X)/2
怎样把下列三角函数转化为y=Asin(ωx+φ)形式
(1)y=sin[2X+(π/8)]cos2X 结果为:y=1/2sin〔4X+(π/8)]+1/2sin(π/8)
(2)y=cos^2 * 4X 结果为:y=(1+cos8X)/2
怎样把下列三角函数转化为y=Asin(ωx+φ)形式(1)y=sin[2X+(π/8)]cos2X 结果为:y=1/2sin〔4X+(π/8)]+1/2sin(π/8)(2)y=cos^2 * 4X 结果为:y=(1+cos8X)/2
(1)
因sin(A+B)+sin(A-B)
=(sinAcosB+cosAsinB)+(sinAcosB-cosAsinB)
=2sinAcosB
所以,sinAcosB=(1/2)[sin(A+B)+sin(A-B)]
上面这个叫做积化和差公式,类似可以得到cosAsinB,cosAcosB,sinAsinB的公式
y=sin[2x+(π/8)]cos2x
=(1/2){sin[4x+(π/8)]+sin(π/8)}
=(1/2)sin[4x+(π/8)]+(1/2)sin(π/8)
(2)
因cos2A
=cos(A+A)
=cosAcosA-sinAsinA
=cos²A-sin²A……(*)
=cos²A-(1-cos²A)
=2cos²A-1……(**)
(*)和(**)是是余弦二倍角公式的两种不同形式
所以cos²A=(1+cos2A)/2
直接代入即可
y=cos²(4x)=(1+cos8x)/2
(1)y=sin[2X+(π/8)]cos2X 结果为:y=1/2sin〔4X+(π/8)]+1/2sin(π/8)
套用公式:2sinαcosβ=sin(α+β)+sin(α-β)就可得出
(2)y=cos^2 * 4X 结果为:y=(1+cos8X)/2
套用公式:cos(α/2)=±√[(1+cosα)/2]就可得出