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我会画y=3cos2x的图像,但是y=3cosπ/2乘以x的图像我就蒙了.图像变化我都会,但是π/2和2不太一样啊.当x前的系数大于1时,是缩短当系数在0到1时,是拉伸.但是π/2怎么搞啊
题目详情
我会画y=3cos2x的图像,但是y=3cos π/2 乘以 x 的图像我就蒙了.
图像变化我都会,但是 π/2 和2 不太一样啊.
当x前的系数大于1 时,是缩短
当系数在 0 到 1时,是拉伸.但是 π/2 怎么搞啊
图像变化我都会,但是 π/2 和2 不太一样啊.
当x前的系数大于1 时,是缩短
当系数在 0 到 1时,是拉伸.但是 π/2 怎么搞啊
▼优质解答
答案和解析
你可以直接代几个 x 的值, 你就会发现它的规则:
x = 0, (pi/2)*x = 0, 3cos[(pi/2)*x] = cos(0) = 3
x = 1, (pi/2)*x = pi/2, 3cos[(pi/2)*x] = cos(pi/2) = 0
x = 2, (pi/2)*x = pi, 3cos[(pi/2)*x] = cos(pi) = -3
x = 3, (pi/2)*x = 3pi/2, 3cos[(pi/2)*x] = cos(3pi/2) = 0
x = 4, (pi/2)*x = 2pi, 3cos[(pi/2)*x] = cos(2pi) = 3
x = 5, (pi/2)*x = 5pi/2, 3cos[(pi/2)*x] = cos(5pi/2) = cos(pi/2) = 0
周期为4.
从理论上看, (pi/2) * 4 = 2pi, 所以 (pi/2) * (x+4) = (pi/2) * x + 2pi
所以 f(x+4) = f(x), 周期为4.
x = 0, (pi/2)*x = 0, 3cos[(pi/2)*x] = cos(0) = 3
x = 1, (pi/2)*x = pi/2, 3cos[(pi/2)*x] = cos(pi/2) = 0
x = 2, (pi/2)*x = pi, 3cos[(pi/2)*x] = cos(pi) = -3
x = 3, (pi/2)*x = 3pi/2, 3cos[(pi/2)*x] = cos(3pi/2) = 0
x = 4, (pi/2)*x = 2pi, 3cos[(pi/2)*x] = cos(2pi) = 3
x = 5, (pi/2)*x = 5pi/2, 3cos[(pi/2)*x] = cos(5pi/2) = cos(pi/2) = 0
周期为4.
从理论上看, (pi/2) * 4 = 2pi, 所以 (pi/2) * (x+4) = (pi/2) * x + 2pi
所以 f(x+4) = f(x), 周期为4.
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