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已知向量m=(根号3sinx/4,1),向量n=(Cosx/4,Cos^2...

、先将f(x)=m*n化成一个函数式子,f(x)=m*n=√3sinxcosx/16+cosx/4,根据二倍角公式,等于三十二分之根号三倍的sin2x加上八分之cos2x加上1,。并进行提取公因式得到十六分之一倍的括号里二分之根号三倍的sin2x加上二分之cos2x括号结束再加上一,...

、先将f(x)=m*n化成一个函数式子,f(x)=m*n=√3sinxcosx/16+cos??x/4,根据二倍角公式,等于三十二分之根号三倍的sin2x加上八分之cos2x加上1,。并进行提取公因式得到十六分之一倍的括号里二分之根号三倍的sin2x加上二分之cos2x括号结束再加上一...

(2a-c)cosB=bcosC 由正弦定理:(2sinA-sinC)cosB=sinBcosC 2sinAcosB-sinCcosB=sinBcosC sinBcosC+sinCcosB=2sinAcosB sin(B+C)=2sinAcosB sinA=2sinAcosB 1=2cosB cosB=1/2 得B=60° f(x)=向量m乘以向量n=根号3sinx/4 cosx/4 +cos^2 x/4=√3/2* s...

m.n=1 (√3sin(x/4),1). (cos(x/4),(cos(x/4))^2)=1 √3sin(x/4). (cos(x/4)+ (cos(x/4))^2 =1 (√3/2)sin(x/2) +( cos(x/2) +1) /2=1 ((√3/2)sin(x/2)+(1/2)cosx/2) = 1/2 cos(π/3 -x/2) = 1/2 [cos(π/3 -x/2)]^2 = 1/4 (cos(2π/3 -x) +1)/2 = 1/4...

第一问,m*n=(2√3sinx/4,2)*(cosx/4,cos^2x/4)=2√3sinx/4*cosx/4+2cos^2x/4=√3sinx/2+(1+cosx/2)=1+2(√3/2*sinx/2+1/2cosx/2)=1+2sin(x/2+π/6)=1, 所以,sin(x/2+π/6)=1/2,那么x/2+π/6=π/4+kπ,其中k为整数,则x+π/3=π/2+2kπ, 故cos(x+π/3)=...

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f(x)=向量m.向量n=√3sin(x/4)cos(x/4)-[cos(x/4)]^2 =√3/2.sin(x/2)-1/2-1/2.cos(x/2) =sin(x/2-π/6)-1/2 令t=x/2-π/6∈[2kπ-π/2,2kπ+π/2] x∈[4kπ-2π/3,4kπ+4π/3] 此为f(x)的增区间。

m*n =√3sin(x/4)cos(x/4) cos²(x/4) =(√3/2)sin(x/2) ( cos(x/2) 1) /2 =((√3/2)sin(x/2) (1/2)cosx/2) 1/2 =cos(π/3 -x/2) 1/2 1)∵m*n=1 ∴f(x)=cos(π/3-x/2) 1/2=1 cos(π/3 -x/2) = 1/2 cos(2π/3 -x) =2cos²(π/3-x/2)-1=1/2

(1) m*n=2√3sinx/4cosx/4+2cos^2x/4 =√3sinx/2+2cos^2x/4-1+1 =√3sinx/2+cosx/2+1 =2sin(x/2+π/6)+1 2sin(x/2+π/6)+1=2 sin(x/2+π/6)=1/2 cos(x+π/3)=cos2(x/2+π/6)=1-2sin^2(x/2+π/6)=1/2 (2) f(x)的取值范围[-1,3]

f(x)=根号3*sinx/4*cosx/4+cos^2*x/4+1/2 =((根号3)/2)sinx/2+(cosx/2+1)/2+1/2 =((根号3)/2)sinx/2+(1/2)cosx/2+1 =cos30°sinx/2+sin30°cosx/2+1 =sin(x/2+30°)+1 周期为4π 对称中心为(-π/3, 1)

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