Question

求解高一三角函数题

Answer 1

f(x) = (1/2)*2sinxcosx + 1 - cos²x = (1/2)sin2x + 1 - (1/2)(cos2x + 1)
= (1/2) + (1/2)(sin2x - cos2x) = 1/2 + (1/√2)[(1/√2)sin2x - (1/√2)cos2x]
= 1/2 + (1/√2)[sin2xcos(π/4) - cos2xsin(π/4)]
= (1/√2)sin(2x - π/4) + 1/2
(1)周期π
(2)正弦函数的值域为[-1, 1], f(x)的值域为[1/2-1/√2, 1/2+1/√2], 即[(1 - √2)/2, (1 + √2)/2]
(3)x在[0, π/4]内, 则2x - π/4在[-π/4, π/4]上，f(x)此时为增函数, f(0) = 0, f(π/4) = 1, f(x)的值域为[0, 1]
(4) -π/2 +2kπ< 2x - π/4 < π/2 + 2kπ即-π/8 +kπ< x < 3π/8 + kπ时f(x)为增函数
π/2 +2kπ< 2x - π/4 < 3π/2 + 2kπ即3π/8 +kπ< x < 7π/8 + kπ时f(x)为减函数
这里k为整数

(5)
f(x) = (1/√2)sin[2(x - π/8)] + 1/2
将sinx上每一点的纵坐标不变，横坐标变为原来的1/2得sin2x
再将其向右平移π/8得sin[2(x - π/8)] = sin(2x - π/4)
再将每一点的横坐标不变，纵坐标变为原来的1/√2得 (1/√2)sin[2(x - π/8)]
最后将上图向上平移1/2