cos2x (cos^2sin^2)的积分

cos2x≠02x≠kπ+π/2,k∈Z∴f(x)定义域为{x|x≠kπ/2+π/4,k∈Z}∵对于x∈{x|x≠kπ/2+π/4,k∈Z},都有-x∈{x|x≠kπ/2+π/4,k∈Z}关于原点对称

诱导公式f(x)=(1+2cos²x-1)/(4cosx)+asin(x/2)cos(x/2)=(cosx)/2+a/2*sinx=(a/2)sinx+(1/2)cosx=√[(a/2)&s

cos^4(x)-sin^4(x)=[cos^2(x)-sin^2(x)][cos^2(x)+sin^2(x)](其中[cos^2(x)+sin^2(x)]=1)=[cos^2(x)-sin^2(x)

cos2x/(cosx+sinx)=(cos²x-sin²x)/(cosx+sinx)=(cosx+sinx)(cosx-sinx)/(cosx+sinx)=cosx-sinx求导

=（cos^2x）'cos2xcosx^2cos^3x+cos^2x(cos2x)'cosx^2cos^3x+cos^2xcos2x(cosx^2)'cos^3x+cos^2xcos2xcosx^2(

∫cos2x/(sin²x*cos²x)dx=∫cos2x/(1/2*sin2x)²dx=4∫cos2x/(sin²2x)dx=4∫csc2x*cot2xdx=

3sin^2x+cos^2x=2sin^2x+sin^2x+cos^2x=2sin^2x+1=2+2sin^2x-1=2-cos2x

cos2x+cos(x+π/3)+sin(x+π/6)+3sin^2x=cos2x+1/2*cosx-根号3/2*sina+根号3/2*sinx+1/2*cosx+3sin^2x=cos2x+cosx

1.∫cos(2x)/(cos²xsin²x)dx=∫cos(2x)sec²xcsc²xdx=4∫cot(2x)csc(2x)dx=2∫cot(2x)csc(2

1.将分母变为sin2x即原式为∫[（4cos2x/sin^2（2x））]dx2.进行换元即2x变为t,原式变为∫[（2cos2x/sin^2t）]dt.3继续换元,可观察到（sint）'=cost.

f(x)=cos2x+2sinx=[1-2sin^2(x)]+2sinx=-2sin^2(x)+2sinx+1

cosx+cos2x+...+cosnx=1/2[(cosx+cosnx)+(cos2x+cos(n-1)x)+...+(cosnx+cosx)]=[cos(n+1)x/2][cos((n-1)x/2

1、y=(cos^2x+sin^2x)^2-2cos^2xsin^2x=1-1/2(sin2x)^2=1-1/4(1-cos4x)=3/4+1/4cos4x周期T=2pi/4=pi/22、y=(根3/

即cos^8x-sin^8x=(cos^4x+sin^4x)(cos^4x-sin^4x)=(cos^4x+sin^4x)(cos²x+sin²x)(cos²x-sin&

两边求平方得：sin(x/2)^2+cos(x/2)^2+2sin(x/2)cos(x/2)=1/4-->1+sin(x)=1/4-->sin(x)=-3/4cos(2x)=1-2sin(x)^2=1

sin^4x+cos^4x-2cos2x=(sin^2x+cos^2x)^2-2*sin^2x*cos^2x-2cos2x=1-(sin(2x)^2)/2-2*cos(2x)=3/4+1/4*cos(

sinx(cos^22x-sin^22x)+2cosxcos2xsin2x=sinxcos4x+cosxsin4x=sin(x+4x)=sin5x

因为sin^2(X)+cos^2(X)=1所以原式=1-cos^2（x）-cos^2（x）=1-2cos^2(x)=-(2cos^2(x)-1)=-cos2x

证明：sin(π/4+x)/sin(π/4-x)+cos(π/4+x)/(π/4-x)?最后少了一个三角函数符号,请核对题目后追问.再问：sin(π/4+x)/sin(π/4-x)+cos(π/4+x

sin2x+sin(π-x)/2cos2x+2sin^2x+cos(-x)=sin2x+sin(x)/2cos2x+2sin^2x+cos(x)=(2cosxsinx+sinx)/2cos²