分类：教育/科学-学习帮助
问题：三角(sin10)^2+(cos40)^2+sin10cos40 
(sin10)^2+(c40)^2+sin10cos40
回答：注意 10°+50°+120°=180°,
由正弦定理和余弦定理得
(sin10)^2+(cos40)^2+sin10cos40
=(sin10°)^2+(sin50°)^2+sin10°*sin50°
=(sin10°)^2+(sin50°)^2-2sin10°*sin50°*cos120°
=(sin120°)^2=3/4.


根据恒等式(a^3-b^3)/(a-b)=a^2+b^2+ab
及三倍角公式sin3t=3sint-4(sint)^3得.
(sin10)^2+(cos40)^2+sin10cos40
=(sin10°)^2+(sin50°)^2+sin10°*sin50°
=[(sin50°)^3-(sin10°)^3]/(sin50°-sin10°)
=(3sin50°-sin150°-3sin10°+sin30°)/[4(sin50°-sin10°)]
=(3sin50°-3sin10°)/[4(sin50°-sin10°)]
=3/4.