sin的4次方8分之π cos的4次方8分之π

(sina)^6+(cosa)^6=((sina)^2)^3+((cosa)^2)^3=((sina)^2+(cosa)^2)[((sina)^2)^2-(sina)^2(cosa)^2+((cosa

(1)因为sinx+sin^2x=1又因为sin^2x+cos^2x=1两式相减得:sinx-cos^2x=0sinx=cos^2x则:cos^2x+cos^6x+cos^8x=cos^2x+(cos

sin^6A+cos^6A=(sin^2A+cos^2A)(sin^4A-sin^2A*cos^2A+cos^4A)=sin^4A-sin^2A*cos^2A+cos^4A=sin^4A+2sin^2

y=(sinx)^4+(cosx)^4=[(sinx)^2+(cosx)^2]^2-2(sinx)^2(cosx)^2=1-(1/2)[sin(2x)]^2=1-(1/4)2[sin(2x)]^2=1

题目说的不清楚啊最好弄张图片上来

(sinA)^4-(sinA)^2+(cosA)^2=[1-(cosA)^2]^2+2(cosA)^2-1展开,化简得到等于(cosA)^4,就是cosA的四次方

(sin^2x+cos^2x)^2=1sin的4次方x＋cos的4次方x+2sin的2次方x乘cos的2次方x=1-2sin的2次方x乘cos的2次方xsin的4次方x＋cos的4次方x=1-2sin

⑴、原式=[(-sina)^2*(-cosa)]/[tana*tana*(-cosa)^3]=sin^2a/(tan^2a*cos^2a)=sin^2a/sin^2a=1；⑵、原式=[(-sina)*

(sina)^4+(sina)^2(cosa)^2+(cosa)^2=(sina)^2[(sina)^2+(cosa)^2]+(cosa)^2=(sina)62+(cosa)^2=1

sin^4a+cos^4a=1(sin^2a+cos^2a)^2-2sin^2acos^2a=1sinacosa=0(sina+cosa)^2=sin^2a+cos^2a+2sinacosa=1+0=

"sin(β加π)等于负的13分之20"有误?cos(α加β)=cosαcosβ-sinαsinβ

sin的4次方α-cos的4次方α=(sin的2次方α+cos的2次方α)(sin的2次方α-cos的2次方α)=sin的2次方α-cos的2次方α=2sin的2次方α-1=2*1/5-1=-3/5

y=(cos²πx+sin²πx)(cos²πx-sin²πx)=1*cos2πx=cos2πx所以T=2π/2π=1

3/5.再问：详细过程，谢谢。再答：cos[sin-1(4/5)]=cos[arcsin(4/5)]=±根下（1-sin^2[arcsin(4/5)]=±根下(1-(4/5)^2)=±3/5

化简后发现右式与左式为相反数,即右式=-√2/4sin(a+b)=sinacosb+cosasinbsin(a-b)=sinacosb-cosasinbcos(a+b)=cosacosb-sinasi

cos的4次方12分之派-sin的4次方12分之派=(cos²π/12+sin²π/12)(cos²π/12-sin²π/12)=1·（cos²π/1

sinα+cosα=√21+2sinαcosα=2sin2α=12α=2kπ+π/2α=kπ+π/4sin的4次方α-cos的4次方α=0

用数学归纳法可证明n次方等于cosnθsinnθ-sinnθcosnθ

1.sinx的4次方+cosx的4次方=（sinx的平方+cosx的平方）^2-2*sinx的平方*cosx的平方=1-1/2(sin2x的平方)=1/4cos4x+3/4T=Pi/22.[Pi/4+

(cosπ/8+sinπ/8)(cos^3π/8+sin^3π/8)=[(cosπ/8+sinπ/8)^2](cos^2π/8-cosπ/8sinπ/8+sin^2π/8)=(1+2cosπ/8sin