第1个回答 2013-03-03
f(x)=ab=(cosx+sinx,sinx)*(cosx-sinx,2cosx)
=cos²x-sin²x+2sinxcosx=sin2x+cos2x
=√2sin(2x+π/4)
(1)周期为:T=π
(2)-π/4≤x≤π/4, -π/2≤2x≤π/2
- π/4=π/4 -π/2≤x+π/4≤π/2+π/4=3π/4
- π/4≤2x+π/4≤3π/4
-√2/2≤sin(2x+π/4)≤1
-1≤f(x)=√2sin(2x+π/4)≤√2
所以值域为:f(x)∈[-1,√2]
第2个回答 2013-03-03
f(X)=ab=(cosx+sinx)(cosx-sinx)+2sinxcosx
=cos^2x-sin^2x+2sinxcosx
=cos2x+sin2x
=√2/2sin(2x+π/4)
T=2π/2=π
x在[-π/4,π/4)]
2x+π/4在[-π/4,3π/4]
当x=π/8时有最大值=√2/2
当x=-π4/时有最小值=-√2/2
第3个回答 2013-03-03
解:f(x)=(cosx+sinx)*(cosx-sinx)+sinx*2cosx
=cos²x-sin²x+2sinx*cosx=cos2x+sin2x=√2sin(2x+π/4)
①最小正周期为π;
②x∈[-π/4,π/4],2x+π/4∈[-π/4,3π/4],最小值为-1,最大值为√2
第4个回答 2013-03-03
解:f(x)=(cosx+sinx)*(cosx-sinx)+2sinxcosx=cos²x-sin²x+sin2x=cos2x+sin2x
=√2(√2/2*sin2x+√2/2cos2x)=√2(cosπ/4*sin2x+sinπ/4*cos2x)=√2sin(2x+π/4)
所以(1)f(x)的最小正周期是π
(2)x∈[-π/4,π/4]所以2x+π.4∈[-π/4,3π/4]令t=2x+π.4,则f(t)=√2sint,t∈[-π/4,3π/4]
可以得到f(t)的最大值是f(π/2)=√2*1=√2,最小值为f(-π/4)=√2*(-√2/2)=-1