Question

三角函数计算

sin(x+θ)-sinx=2sinθ/2cos(x+θ/2)
详细写一些过程

Answer 1

你好！

这里用到公式 sina - sinb = 2sin[(a-b)/2]cos[(a+b)/2]

推导如下：
sin(x+y) = sinxcosy+cosxsiny
sin(x-y)=sinxcosy - cosxsiny
两式相减 sin(x+y) - sin(x-y) = 2 cosxsiny
令a=x+y b=x-y
则 x= (a+b)/2 y=(a-b)/2
∴sina - sinb = 2sin[(a-b)/2]cos[(a+b)/2]

令a=x+θ b=x 代入上式即得
sin(x+θ)-sinx=2sinθ/2cos(x+θ/2)

Answer 2

sin(x+θ)-sinx=sin[(x+θ/2)+θ/2]-sin[(x+θ/2)-θ/2]=sin(x+θ/2)*cos(θ/2)+cos(x+θ/2)*sin(θ/2)-[sin(x+θ/2)*cos(θ/2)-cos(x+θ/2)*sin(θ/2)]=2sinθ/2cos(x+θ/2)

Answer 3

你这等式输正确着没

追问

你懂不懂的，shift！！！