作业帮已知函数f(x)=cos(x+x/6)-sin(x-2π/3)+sinx+a的最大值为1.求常数a的值?求使f(x)≥0
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已知函数f(x)=cos(x+x/6)-sin(x-2π/3)+sinx+a的最大值为1.求常数a的值?求使f(x)≥0成立的x的取值范围
f(x) = cos(x+π/6)-sin(x-2π/3)+sinx+a
= cos(x+π/6)-{-sin(π+x-2π/3)} + sinx + a
= cos(x+π/6)+sin(x+π/3)} + sinx + a
= cos(x+π/6)+cos{π/2-(x+π/3)} + sinx + a
= cos(x+π/6)+cos{π/6-x} + sinx + a
= cos(x+π/6)+cos{x-π/6} + sinx + a
= cosxcosπ/6-sinxsinπ/6 + cosxcosπ/6+sinxsinπ/6 + sinx + a
= 2cosxcosπ/6 + sinx + a
= 2(cosxcosπ/6 + sinxsinπ/6) + a
= 2cos(x-π/6)+a
-2+a≤2cos(x-π/6)+a≤2+a
2+a=1
a=-1
= cos(x+π/6)-{-sin(π+x-2π/3)} + sinx + a
= cos(x+π/6)+sin(x+π/3)} + sinx + a
= cos(x+π/6)+cos{π/2-(x+π/3)} + sinx + a
= cos(x+π/6)+cos{π/6-x} + sinx + a
= cos(x+π/6)+cos{x-π/6} + sinx + a
= cosxcosπ/6-sinxsinπ/6 + cosxcosπ/6+sinxsinπ/6 + sinx + a
= 2cosxcosπ/6 + sinx + a
= 2(cosxcosπ/6 + sinxsinπ/6) + a
= 2cos(x-π/6)+a
-2+a≤2cos(x-π/6)+a≤2+a
2+a=1
a=-1
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