數學問題一問..(thx)萬分感激!!!!!

1)sin40`sin80`sin160`
= sin40 sin80 sin20
= (-1/2)[cos(40+20) - cos(40-20)]sin80
= (-1/2)(1/2 - cos20) sin80
= (-1/4)sin80 + (1/2)cos20 sin80
= (-1/4)sin80 + (1/2)(1/2)[sin(20+80) - sin(20-80)]
= (-1/4)sin80 + (1/4)(sin100 + sin60)
= (-1/4)sin80 + (1/4)(sin80 + sin60)
= (1/4)sin60
= √3 / 8

2)sin20`cos70`+sin10`sin50
= (1/2)[sin(20+70) + sin(20-70)] + (-1/2)[cos(10+50) - cos(10-50)]
= (1/2)(sin90 - sin50) - (1/2)(cos60 - cos40)
= (1/2)(1) - (1/2)sin50 - (1/2)(1/2) + (1/2)(sin50)
= 1/2 - 1/4
= 1/4