一道高一必修四三角函数题“π”这个是“派”印刷体好像看着不像麻烦写下步骤,晚上自习课一直在看这道题.求求你了,我数学学的不好,想弄懂啊!已知函数f（x）=Sin（π-ωx）Cosωx+Cos²ωx,（ω＞0）的最小正周期为π.（1）求ω的值（2）将函数y=f（x）的图像上各点的横坐标缩短到原来的1/2倍,纵坐标不变,得到函数y=g（x）的图像,求函数g（x）在区间[0,π/16]上的最小值.（3）对于f（x）若x∈[-π/4,π/4],则f（x）的值域为多少?（4）对于g（x）若g（x）∈[0,
2019-05-27
一道高一必修四三角函数题
“π”这个是“派”印刷体好像看着不像
麻烦写下步骤,晚上自习课一直在看这道题.求求你了,我数学学的不好,想弄懂啊!
已知函数f（x）=Sin（π-ωx）Cosωx+Cos²ωx,（ω＞0）的最小正周期为π.
（1）求ω的值
（2）将函数y=f（x）的图像上各点的横坐标缩短到原来的1/2倍,纵坐标不变,得到函数y=g（x）的图像,求函数g（x）在区间[0,π/16]上的最小值.
（3）对于f（x）若x∈[-π/4,π/4],则f（x）的值域为多少?
（4）对于g（x）若g（x）∈[0,1],则x的取值集合为多少?
（5）对于f（x）,求其单调增区间.
（6）对于g（x）求其对称中心和对称轴.
优质解答
f（x）=Sin（π-ωx）Cosωx+Cos²ωx
=sinωxcosωx+1/2(1+cos2ωx)
=1/2sin2ωx+1/2cos2ωx+1/2
=√2/2sin(2ωx+π/4)+1/2
T=2π/2ω=π,ω=1,
f(x)=√2/2sin(2x+π/4)+1/2
(2)g(x)=√2/2sin[2*(2x+π/4)]+1/2
=√2/2sin(4x+π/2)+1/2
=√2/2cos4x+1/2
0≤x≤π/16
0≤4x≤π/4
g(x)min=g(π/16)=1
（3） f(x)=√2/2sin(2x+π/4)+1/2
-π/4≤x≤π/4
-π/4≤2x+π/4≤3π/4
f(-π/4)≤ f(x)≤f(π/8)
0≤ f(x)≤√2/2+1/2
(4)g(x)= √2/2cos4x+1/2
0 ≤√2/2cos4x+1/2 ≤1
-1/2 ≤√2/2cos4x ≤1/2
-√2/2 ≤cos4x ≤√2/2
2kπ+π/4 ≤4x ≤2kπ+3π/4,2kπ+5π/4 ≤4x ≤2kπ+7π/4
kπ/2+π/16 ≤x ≤kπ/2+3π/16,kπ/2+5π/16 ≤x ≤2kπ+7π/16
(5) f(x)=√2/2sin(2x+π/4)+1/2
2kπ-π/2 ≤2x+π/4 ≤2kπ+π/2
kπ-3π/8≤x≤kπ+π/8
(6)g(x)= √2/2cos4x+1/2
.对称中心(kπ/4+π/8,1/2)
对称轴x=kπ/4
f（x）=Sin（π-ωx）Cosωx+Cos²ωx
=sinωxcosωx+1/2(1+cos2ωx)
=1/2sin2ωx+1/2cos2ωx+1/2
=√2/2sin(2ωx+π/4)+1/2
T=2π/2ω=π,ω=1,
f(x)=√2/2sin(2x+π/4)+1/2
(2)g(x)=√2/2sin[2*(2x+π/4)]+1/2
=√2/2sin(4x+π/2)+1/2
=√2/2cos4x+1/2
0≤x≤π/16
0≤4x≤π/4
g(x)min=g(π/16)=1
（3） f(x)=√2/2sin(2x+π/4)+1/2
-π/4≤x≤π/4
-π/4≤2x+π/4≤3π/4
f(-π/4)≤ f(x)≤f(π/8)
0≤ f(x)≤√2/2+1/2
(4)g(x)= √2/2cos4x+1/2
0 ≤√2/2cos4x+1/2 ≤1
-1/2 ≤√2/2cos4x ≤1/2
-√2/2 ≤cos4x ≤√2/2
2kπ+π/4 ≤4x ≤2kπ+3π/4,2kπ+5π/4 ≤4x ≤2kπ+7π/4
kπ/2+π/16 ≤x ≤kπ/2+3π/16,kπ/2+5π/16 ≤x ≤2kπ+7π/16
(5) f(x)=√2/2sin(2x+π/4)+1/2
2kπ-π/2 ≤2x+π/4 ≤2kπ+π/2
kπ-3π/8≤x≤kπ+π/8
(6)g(x)= √2/2cos4x+1/2
.对称中心(kπ/4+π/8,1/2)
对称轴x=kπ/4