题目
f(x)=sinx^2+sinxcosx,求f(π/4),求x属于[0,π/2],求f(x)最大值及相应x值
提问时间:2021-10-28
答案
f(x)=sinx^2+sinxcosx
=(1-cos2x)/2+1/2sin2x
=(1-cos2x)/2+1/2sin2x
=1/2sin2x-1/2cos2x+1/2
=√2/2*(√2/2*sin2x-√2/2*cos2x)+1/2
=√2/2*(sin2xcosπ/4-cos2xsinπ/4)+1/2
=√2/2*sin(2x-π/4)+1/2
f(π/4)=√2/2*sin(2*π/4-π/4)+1/2
=√2/2*sin(π/4)+1/2
=√2/2*√2/2+1/2
=1/2+1/2
=1
x∈[0,π/2]
2x∈[0,π]
2x-π/4∈[-π/4,3π/4]
f(x)最大值为:
√2/2*sin(2x-π/4)+1/2
=√2/2+1/2
2x-π/4=π/2
2x=3π/4
x=3π/8
=(1-cos2x)/2+1/2sin2x
=(1-cos2x)/2+1/2sin2x
=1/2sin2x-1/2cos2x+1/2
=√2/2*(√2/2*sin2x-√2/2*cos2x)+1/2
=√2/2*(sin2xcosπ/4-cos2xsinπ/4)+1/2
=√2/2*sin(2x-π/4)+1/2
f(π/4)=√2/2*sin(2*π/4-π/4)+1/2
=√2/2*sin(π/4)+1/2
=√2/2*√2/2+1/2
=1/2+1/2
=1
x∈[0,π/2]
2x∈[0,π]
2x-π/4∈[-π/4,3π/4]
f(x)最大值为:
√2/2*sin(2x-π/4)+1/2
=√2/2+1/2
2x-π/4=π/2
2x=3π/4
x=3π/8
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