求めます：cos^8 x-sin^8 x-cos 2 x=1/8(cos 6 x-cos 2 x) また、csc^4 a(1-cos^4 a)-2 cot^2 a

求めます：cos^8 x-sin^8 x-cos 2 x=1/8(cos 6 x-cos 2 x) また、csc^4 a(1-cos^4 a)-2 cot^2 a

cos^8 x-sin^8 x-cos 2 x
=(cos^4 x+sin^4 x)(cos^2 x+sin^2 x)(cos^2 x-sin^2 x)-cos 2 x
=(cos^4 x+sin^4 x)*1*cos 2 x-cos 2 x
=[(cos^2 x+sin^2 x)^2-2(sinxcox)^2]cos 2 x-cos 2 x
=[1-2(sinxcox)^2]cos 2 x-cos 2 x
=[1-2(sinxcox)^2-1]cos 2 x
=-(2 sinxcox)^2/2*cos 2 x
=-sin^2 2 x/2*co 2 x
=-(1+cos 4 x)*cos 2 x/4
1/8(cos 6 x-cos 2 x)
=1/8*(-2)*sin(8 x/2)sin(4 x/2)
=-1/4 sin 4 xsin 2 x
=-1/4*(2 sin 2 xcos 2 x)sin 2 x
=-1/4*2(sin 2 x)^2 cos 2 x
=-1/4(1+cos 4 x)cos 2 x
したがって、cos^8 x-sin^8 x-cos 2 x=1/8(cos 6 x-cos 2 x)

求めます：cos^8(x)-sin^8(x)+(1/4)sin 2 xsin 4 x=cos 2 x 注：冒頭のcos、sinの8は8乗で、xは8上ではなく、cos、sinの上にあります。

cos^8(x)-sin^8(x)+(1/4)sin 2 xsin 4 x=[cos^4(x)+sin^4(x)*[cos^4(x))*[cos^2]+(1/4)*sin 2 x*2 x 2 x 2 x 2 x=cos 2 x=[cos^4(x)*^2****^2***^2***********************************^2 2 2 x x x x x x x^2**^2*********************^2**************+sin^2(...

cos 2 x=3/5 sin^4+cos^4 x=？

(sinx)^4+(cosx)^4
=[(sinx)^2+(cosx)^2]^2-2(sinx)^2(cosx)^2
=1-(1/2)[sin(2 x)]^2
=1-(1/2)[1-cos(2 x)^2]
=1-(1/2)[1-(3/5)^2]
=1-8/25
=17/25

2 x=3/5をcosするなら、sin^4+cos^4 x=いくらですか？

(sinx)^4+(cosx)^4
=[(sinx)^2+(cosx)^2]^2-2(sinx)^2(cosx)^2
=1-(1/2)[sin(2 x)]^2
=1-(1/2)[1-cos(2 x)^2]
=1-(1/2)[1-(3/5)^2]
=1-8/25
=17/25

cos 2 X=ルート2/3をすでに知っていますが、sin^4 X+cos^4の値は？

y=sin^4 x+cos^4 x
=sin^4 x+2 sin^2 xcos^2 x+cos^4 x-2 sin^2 xcos^2 x
=(sin^2 2 x+cos^2 2 x)^2-(1/2)(2 sinxcox)^2
=1-(1/2)(sin 2 x)^2
=1-(1/2)(1-cos^2 2 x)
=1-1/2*(1-2/9)
=1-7/18
=11/18

cox^2=1+cosx/2は1+cos 2 x/2ですか？sin 2 x=1-cosxもありますか？

一つ目の式は（1+cos 2 x）/2
sin 2 x=1 coxは間違っています。sin 2 x=2 sinxcoxです。

証明書を求めます：tan(3 x/2)-tan(x/2)=2 sinx/(cos x+cos 2 x)

tan(3 x/2)-tan(x/2)
=sin(3 x/2)/cos(3 x/2)-sin(x/2)/cos(x/2)(通分)
=[sin(3 x/2)cos(x/2)-cos(3 x/2)sin(x/2)/[cos(3 x/2)cos(x/2)]
=sin(3 x/2-x/2)/[(1/2)(cos 2 x+cox)(積化と差)
=2 sinx/(cox+cos 2 x)
旧式が成立する

証明：tan[3 x/2]-tan[x/2]=2 sinx/[cox+cos 2 x]

tan(3 x/2)-tan(x/2)=sin(3 x/2)/cos(3 x/2)-sin(x/2)/cos(x/2)(x/2)/cos(x/2)=[sin(3 x/2)-cos(3 x/2)(3 x/2)sin(x/2))/[cos(3 x/3))))/[cos(3 x(3 x 3)))))))))/[cos(3 x(3 x(3)))))))/(3)))))))))/[cos(3 x(3 x(3 x(3))))))))))))))))))))))))/2x)故原式成立…

tan(3 x/2)-tan(x/2)-2 sinx/(cox+cos(2 x)=？

tan(3 x/2)-tan(x/2)-2 sinx/(cox+cos 2 x)
=sin(3 x/2)/cos(3 x/2)-sin(x/2)/cos(x/2)-2 sinx/(cox+cos 2 x)
=[sin(3 x/2)cos(x/2)-cos(3 x/2)sin(x/2)/cos(3 x/2)cos(3 x/2)-2 sinx/(cox+cos 2 x)
=2 sin(3 x/2 x/2)/[cos(3 x/2+x/2)+cos(3 x/2 x/2)]-2 sinx/(cox+cos 2 x)
=2 sinx/(cos 2 x+cox)-2 sinx/[cox+cos(2 x)]
=0

sin 2 x/(1-cos 2 x)-(1/tanx)化簡略

sin 2 x/(1-cos 2 x)-(1/tanx)
=2 sinxcox/(2 sin²x)-cosx/sinx
=cox/sinx-cox/sinx
=0