√1-cos²4

1-sin^4a-cos^4a/cos^2a-cos^4a=[1-(sin²a+cos²a)+2sin²acos²a]/cos²a(1-cos

用和差化积公式和分子有理化技巧：an=cos√（n+1）-cos√n=-2sin{[√（n+1）+√n]/2}sin{[√（n+1）-√n]/2}=-2sin{[√（n+1）+√n]/2}sin{1/

g(x)=(sinx)^2(1-cosx)/|sinx|+(cosx)^2(1-sinx)/|cosx|=2cosxsinx-sinx-cosx（x∈（π,5π/4））令sinx+cosx=t（-根号

证明：输入过于麻烦,用换元法吧设A=sin²A,B=sin²B∵sin^4a/sin^2b+cos^4a/cos^2b=1即A²/B+(1-A)²/(1-B)=

cos(α+β)=1/3,cos(α-β)=1/4,求cosα-cosβcos(α+β)=cosαcosβ-sinαsinβ=1/3.(1)cos(α-β)=cosαcosβ+sinαsinβ=1/4

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

3)9/64Half-Angleformula:cos(x)=2cos(x/2)^2-1Inthisquestion:cos(x/2)^2=5/16So:cos(x)=2(5/16)-1=-3/8So

α是锐角,√Cos^2α-4Cosα+4-|（1-Cosα）|=cosa-4cosa+4-1+cosa=-4cosa+3

分子（1+sinα+cosα）（sinα／2-cosα／2）=[2（cosα/2）^2+2sinα/2*cosα/2]（sinα／2-cosα／2）=2cosα/2(sinα／2+cosα／2)(si

1+sinθ+cosθ=2(cosθ/2)^2+2sin(θ/2)cos(θ/2)=2cos(θ/2)[cos(θ/2)+sin(θ/2)]∴分子=2cos(θ/2){[sin(θ/2)]^2-[co

（sinx）^4+（cosx）^4=1即(sinx)^4+(cosx)^4+2(sinx)^2(cosx)^2-2(sinx)^2(cosx)^2=1即[(sinx)^2+(cosx)^2]^2-2(

sinθ+sin^2θ=1,sin^2θ+cos^2θ=1∴sinθ=cos^2θ等式两边同时乘以sinθ有sin^2θ+sin^3θ=sinθ∴sin^3θ=sinθ-sin^2θ=cos^2θ-s

1-sin2a-cos^2(a-π/4)=sin^2(a-π/4)-sin2a=(根号2sina-根号2cosa)^2-sin2a=2(cosa^2-2sinacosa+cosa^2)-sin2a=2

=[1-(sin²a+cos²)(sin^4a-sin²acos²a+cos^4a)]/cos²a(1-cos²a)=[1-(sin^4a+

sinθ-cosθ=-1/5两边平方得1-2sinθcosθ=1/25sinθcosθ=24/50=12/25sin²θ+cos²θ=1两边平方得sinθ^4+cosθ^4=1-2

=sin4Asin2A+cos4Acos2A-cos2A(cos4Acos2A-sin4Asin2A)=cos2A+cos2Acos6A=cos2A(1+cos6A)

解tan2θ=(2tanθ)/(1-tan^2θ)=3/4即3(1-tan^2θ)=8tanθ即3tan^2θ+8tanθ-3=0(3tanθ-1)(tanθ+3)=0∵π/2

[cos^4(π/3+α)]-[cos(π/6-α)]^2=[cos(π/3+α)]^4-[sin(π/3+α)]^4=[cos(π/3+α)]^2-[sin(π/3+α)]^2=cos(2π/3+2

[2cosθ/√(1-sin²θ)]+[√(1-cos²θ)/sinθ]=(2cosθ/|cosθ|)+(|sinθ|/sinθ)cosθ>0,sinθ>0时[2cosθ/√(1-

题目还有问题,应该是“cos²α+cos²β+√2cosα×cosβ=1/2”丢了根号2吧?cos²α+cos²β+cosαcosβ=(1+cos2α)/2+(