Question

急！高一数学，会的请进！

cos47/6π=
sin8/7π=
1+sin（x-2π）乘sin（π+x）-2cos平方（-x)=
sin25/6π+cos25/3π+tan(-25/4π）=

Answer 1

cos47/6π=cos(8π-1/6π)=-cosπ/6=-√3/2

sin8/7π=sin(π+π/7)=-sinπ/7

1+sin（x-2π）乘sin（π+x）-2cos平方（-x)
=1+sinx*(-sinx）-2cos^2 x
=1-sin^2 x-2cos^2 x
=cos^2 x-2cos^2 x
=-cos^2 x

sin25/6π+cos25/3π+tan(-25/4π）
=sin(4π+π/6)+cos(8π+π/3)+tan(-6π-π/4）
=sinπ/6+cosπ/3+tan(-π/4）
=sinπ/6+cosπ/3-tanπ/4
=1/2+1/2-1
=0

Answer 2

cos47/6π=根号3/2
sin8/7π=-0.434
1+sin（x-2π）乘sin（π+x）-2cos平方（-x)=0

sin25/6π+cos25/3π+tan(-25/4π）= 0

Answer 3

1、cos47/6π=cos(-π/6)=根号3/2

2、sin8/7π=-sinπ/7

3、1+sin（x-2π）乘sin（π+x）-2cos平方（-x)
=1-sin^2x-2cos^2x
=-sin^2x+sin^2x-cos^2x
=-cos^2x

4、sin25/6π+cos25/3π+tan(-25/4π）
=sinπ/6+cos1/3π+tan(-π/4）
=0.5+0.5-1
=0

Answer 4

cos47/6π=cos(8π - π/6)=cos(π/6)=(√3)/2

sin8/7π=sin(π + π/7)=-sin(π/7)

原式=1 + sin[-(2π-x)](-sinx) - 2(cosx)^2
=1 + sinx(-sinx) - 2(cosx)^2
=[(sinx)^2 + (cosx)^2] - (sinx)^2 - 2(cosx)^2
=-(cosx)^2

原式=sin(4π + π/6) + cos(8π + π/3) - tan(6π + π/4)
=sin(π/6) + cos(π/3) - tan(π/4)
=1/2 + 1/2 - 1
=0