急求三角函数cos10°cos50°cos70°的值.
问题描述：
急求三角函数cos10°cos50°cos70°的值.
答
cos10°cos50°cos70°
=1/2*{cos(10°+50°)+cos(10°-50°)}*cos70°
=1/2(cos60°+cos40°)cos70°
=1/2(1/2+cos40°)cos70°
=1/4*cos70°+1/2*cos40°cos70°
=1/4*cos70°+1/2*1/2{cos(40°+70°)+cos(40°-70°)}
=1/4*cos70°+1/4[cos110°+cos30°]
=1/4(cos70°+cos110°+cos30°)
=1/4[cos(180-110°)+cos110°+cos30°]
=1/4[-cos110°+cos110°+cos30°]
=1/4*cos30°
=1/4*√3/2
=√3/8