一道高一必修四三角函数题
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发布时间：2022-08-16 21:23
共3个回答
热心网友
时间：2023-10-09 02:16
=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
热心网友
时间：2023-10-09 02:16
f(x)=sin(π-ωx)cosωx+cos²ωx
f(x)=sin(ωx)cosωx+cos²ωx
f(x)=(1/2)sin(2ωx)+[1+cos(2ωx)]/2
f(x)=(1/2)sin(2ωx)+(1/2)cos(2ωx)+1/2
f(x)=(√2/2)[(√2/2)sin(2ωx)+(√2/2)cos(2ωx)]+1/2
f(x)=(√2/2)[cos(π/4)sin(2ωx)+sin(π/4)cos(2ωx)]+1/2
f(x)=(√2/2)sin(2ωx+π/4)+1/2
T=2π/(2ω)=π
解得：ω=1
解2：
g(x)=(√2/2)sin(4x+π/2)+1/2
g(x)=(√2/2)cos(4x)+1/2
g'(x)=-(2√2)sin(4x)
令：g'(x)≤0，即：-sin(4x)≤0
解得：0≤x≤π/4
即：g(x)的减区间是x∈[0，π/4]
而：[0，π/16]∈[0，π/4]
所以：x∈[0，π/16]时，g(x)是减函数
也就是x=π/16时，g(x)有最小值。
该最小值是：g(x)最小=(√2/2)cos(π/4)+1/2=1
解3：
f(x)=(√2/2)sin(2x+π/4)+1/2
f'(x)=(√2)cos(2x+π/4)
令：f'(x)≥0，即：cos(2x+π/4)≥0
有：2kπ-π/2≤2x+π/4≤2kπ+π/2
解得：kπ-3π/8≤x≤kπ+π/8
f(x)的增区间是x∈[kπ-3π/8，kπ+π/8]
由于[-π/4，π/8]∈[-3π/8，π/8]，所以f(x)在[-π/4，π/8]是增函数，
因此：f(x)∈[f(-π/4)，f(π/8)]
f(-π/4)=(√2/2)sin[2(-π/4)+π/4]+1/2=0
f(π/8)=(√2/2)sin[2(π/8)+π/4]+1/2=(1+√2)/2
即：f(x)∈[0，(1+√2)/2]
太多了，今天懒得做了。
等有功夫再接着做吧。
热心网友
时间：2023-10-09 02:17
=sinωxcosωx+(1+cos2ωx)/2
=1/2sin2ωx+1/2cos2ωx+1/2
=√2/2sin(2ωx+π/4)+1/2
T=2π/2ω=π, ω=1,
所以f(x)=√2/2sin(2x+π/4)+1/2
(2)g(x)=√2/2sin（2*2x+π/4)+1/2
=√2/2sin(4x+π/4)+1/2
0≤x≤π/16
π/4≤4x+π/4≤π/2
所以g(x)min=g(0)=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)=√2/2sin(4x+π/4)+1/2
0 ≤√2/2sin(4x+π/4)+1/2≤1
-1/2 ≤√2/2sin(4x+π/4)≤1/2
-√2/2 ≤sin(4x+π/4)+ ≤√2/2
2kπ-π/4 ≤4x +π/4≤2kπ+π/4, kπ/2-π/8 ≤x ≤kπ/2
(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) =√2/2sin(4x+π/4)+1/2
无对称中心， 对称轴x=kπ/4+π/16
