1.证明sin 2 nx = sin((2n + 1)x) cos x – cos((2n + 1)x) sin x
5个回答
sin[(2n + 1)x]cosx - cos[(2n + 1)x]sinx
= sin[(2n + 1)x - x]
= sin(2nx + x - x)
= sin(2nx)
公式：sin(x ± y) = sinxcosy ± cosxsiny
sin4x/(2sinx) = 1/2,0 < x < π
sin4x = sinx
sin4x - sinx = 0
2cos[(4x + x)/2]sin[(4x - x)/2] = 0,公式sinx - siny = 2cos[(x + y)/2]sin[(x - y)/2]
cos(5x/2)sin(3x/2) = 0
cos(5x/2) = 0 OR sin(3x/2) = 0
5x/2 = π/2 OR 3x/2 = 0 OR 3x/2 = π,∵cos(π/2) = 0,sin(0) = 0,sin(π) = 0
x = π/5 OR x = 0 OR x = 2π/3,已知x > 0所以舍掉x = 0
∴x = π/5 OR x = 2π/3
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