Question

数学题.高一

求值:
1.(sin20度)^2+(cos50度)^2+sin20度*cos50度
2.log2(cos(π/9))+log2(cos(2π/9))+log2(cos(4/9))

见图!

Answer 1

1.(sin20度)^2+(cos50度)^2+sin20度*cos50度
=(1-cos40)/2+(1+cos100)/2+1/2*[sin70-sin30]
=(-cos40+cos100+sin70)/2
=(-cos40-cos80+cos20)/2
=[cos20-cos(60-20)-cos(60+20)]/2
=[cos20-cos60cos20-sin60sin20-cos60cos20+sin60sin20]/2
=[cos20-2cos60cos20]/2
=[cos20-2*1/2*cos20]/2
=0
2.log2(cos(π/9))+log2(cos(2π/9))+log2(cos(4π/9))
=log2[cos(π/9)cos(2π/9)cos(4π/9)]
=log2[cos(π/9)cos(2π/9)cos(4π/9)sin(π/9)/sin(π/9)]
=log2[cos(π/9)sin(π/9)cos(2π/9)cos(4π/9)/sin(π/9)]
=lon2[1/2*sin(2π/9)cos(2π/9)cos(4π/9)/sin(π/9)]
=log2[1/4*sin(4π/9)cos(4π/9)/sin(π/9)]
=log2[1/8*sin(8π/9)/sin(π/9)]
=log2(1/8)
=-3