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设F(sinx)=Cos2x+1,求F(Cosx)

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f(sinx)=1-cos2x =1-(1-2(sinx)²)=2(sinx)² 求 f(x)=2x² f(cosx)=2(cosx)² f(x+1/x)=x²+1/x²=x²+1/x²+2-2=(x+1/x)²-2 f(x)=x²-2 f(x-1/x)=(x-1/x)²-2=x²+1/x²-4

解: (1) f(x)=cos²x-√3sinxcosx+½ =½[1+cos(2x)]-(√3/2)sin(2x)+½ =½cos(2x)-(√3/2)sin(2x)+1 =cos(2x+π/3)+1 最小正周期T=2π/2=π cos(2x+π/3)=1时,f(x)取得最大值f(x)max=1+1=2 cos(2x+π/3)=-1时,f(x)取得最小值f(...

(Ⅰ)f(x)=1+sin2x+cos2x=2sin(2x+π4)+1所以T=2π2=π(Ⅱ)由0≤x≤π2,得π4≤2x+π4≤5π4当2x+π4=π2,即x=π8时,f(x)max=2+1;当2x+π4=5π4,即x=π2时,f(x)min=0

f(x)=a(1+cosx+sinx)+b=2asin(x+π4)+a+b,(1)当a=-1时,由2kπ+π2≤x+π4≤2kπ+32π,得2kπ+π4≤x≤2kπ+54π,∴f(x)的单调增区间为[2kπ+π4,2kπ+54π](k∈Z);(2)∵0≤x≤π,∴π4≤x+π4≤54π,∴-22≤sin(x+π4)≤1,依题意知a≠0,分两种情况考虑:1...

y=f(x)=sin 2 x+sinx·cosx+cos2x =sin 2 x+1/2sin2x+cos2x =3/2sin2x+cos2x y=f(x)=3/2*sin2x+cos2x

f(sinx/2)=cosx+1 =1-2(sin(x/2))^2 +1 =2- 2(sin(x/2))^2 f(x)=2-2x^2 f(cos(x/2)^2=2-2(cos(x/2))^2 =2-(1+cosx) =1-cosx

(1)由题设f(x)=-sin2x+1+cos2x+1=2cos(2x+π4)+2.∵f(x)-1=0,∴2cos(2x+π4)+2=1,∴cos(2x+π4)=?22,则2x+π4=2kπ+34π或2x+π4=2kπ+54π,k∈Z,得x=kπ+π4或x=kπ+π2,k∈Z,∵x∈(0,π),∴x1=π4,x2=π2,∴x1+x2=34π;(2)由函数y=f(x...

(1)∵f(x)=sinxcosx-3cos2x+32+1=12sin2x-32cos2x+1=sin(2x-π3)+1函数f(x)的最小正周期:T=2π2=π令2x-π3=kπ 解得x=kπ2+π6 (k∈Z)∴函数f(x)图象的对称中心为(kπ2+π6,1)(k∈Z)令2x-π3=kπ+π2 解得x=kπ2+5π12 (k∈Z)∴函数f(x)的对...

(1) f(x)=2sinxcosx+ 3 cos2x (2分)= sin2x+ 3 cos2x (4分)= 2sin(2x+ π 3 ) (6分)所以,函数f(x)的最小正周期为π,(7分)由 2x+ π 3 =kπ+ π 2 ,k∈Z,得 x= kπ 2 + π 12 ,k∈Z,所以,函数f(x)图象的对称轴方程为x= kπ 2 + π 12 ...

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