cos(6x)=sin4x sin2x-搜答案-大师作文网

∫(1+cos^2x)/cos^2xdx=∫1/cos^2x+1dx=∫1/cos^2xdx+x=∫1d（tanx)+x=tanx+x+c

f(x)=（6cos^4x-5cos^2x+1）/（2cos^2x-1）=3cos^2x-1,2cos^2x-1≠0cos2x≠02x≠kπ+π/2,k∈Z解出x得定义域f(x)=3cos^2x-1=

f（x）可化为=1/2+根号3*sin(2x+pai/2)(2x+pai/2)=π/2时有最大值1/2+根号3=1/2+根号3*sin(2x+pai/3)然后(2x+pai/3)=pai/2然后sin

f(x)=[cos(派/6)]^2+sin(派/6)cos(派/6)=[(根号3)/2]^2+[(1/2)*(根号3)/2]=(3/4)+(根号3)/4=(3+根号3)/4.

1)f(x)=1+cos(2x+π/3)-(1+cos2x)/2=1/2-sin2x根号3/2最小值1/2-根号3/2最小正周期π2）c带入得sinC=根号3/2C=π/3A=π-B-C=2π/3-a

（cosx）^2-（sinx）^2=cos2x,变换加移项能的到你写的公式

cos(x+π/3)=sin[π/2-(x+π/3)]=sin(π/6-x)=-sin(x-π/6)所以y=sin(3x+π/3)cos(x-π/6)-cos(3x+π/3)sin(x-π/6)=si

f(x)=-√2sin(2x+π/4)+6sinxcosx-2cos²x+1=-√2(sin2xcosπ/4+cos2xsinπ/4)+3sin2x-2×(1+cos2x)/2+1=-√2(

不知道,sorry

∵cos（x-π/6）=-√3/3∴cosx+cos（x-π/3）=cosx+cosxcosπ/3+sinxsinπ/3=cosx+（1/2）cosx+（√3/2）sinx=（3/2）cosx+（√3

这个方程无解取区间[-π,π]-1

∫cos x×arc cos x=?

但是如何作变换呢?怎样才能反函数也成功换成的简单函数式?请回答的具体些!多谢!

f(pi/12)=sin(pi/3)-cos(pi/2)+(2(cos(pi/12))^2-1)+1==sqrt(3)/2-0+cos(pi/6)+1=sqrt(3)/2+1/2+1=3/2+sqrt

cos(5π/6+x)-sin^2(x-π/6）=cos〔π-(π/6-x)〕-〔1-cos^2(x-π/6)〕=-cos(π/6-x)-〔1-cos^2(x-π/6)〕=-cos(π/6-x)-〔1

(1)cos[(pi/3)-x]=cos[(pi/2-pi/6)-x]=cos[pi/2-(pi/6+x)]=sin[(pi/6+x)]=1/3(2)cos[(2pi/3)+x]=cos[(pi-pi

画一个单位园作两个角：一个：x,一个-x可以看出：cosx=OA/R //: 由余弦函数的定义而

解:⑴f(x)=-1/2+sin(π/6-2x)+cos(2x-π/3)+(cosx)^2=-1/2+sinπ/6cos2x-sin2xcosπ/6+cos2xcosπ/3+sin2xsinπ/3+(

Cos x=-1/2

) y=cos(x-y)

1.两边求导得：y'=-sin(x-y)(1-y')解得y'=sin(x-y)/[sin(x-y)-1]2.y'=-e^-xy''=e^-xy'"=-e^-x3.y'"=(e^2x)'"(sinx)+