搜搜做题作业网求教数学题：α=cos10°、β=cos50° 、γ=cos70° 求α^2+β^2+γ^2
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求教数学题：α=cos10°、β=cos50° 、γ=cos70° 求α^2+β^2+γ^2
α^2+β^2+γ^2
=cos^2 10°+cos^2 50°+cos^2 70°
=1/2(cos20°+1)+1/2(cos100°+1)+1/2cos(140°+1) 由cos2a=2cos^2a-1 而来
=1/2(cos20°+cos100°+cos140°+3)
=1/2(2cos(20°+100°)/2cos(20°-100°)/2+cos(180°-40°)+3)
=1/2(2cos60°cos40-cos40°+3)
=1/2(cos40°-cos40°+3)
=3/2
=cos^2 10°+cos^2 50°+cos^2 70°
=1/2(cos20°+1)+1/2(cos100°+1)+1/2cos(140°+1) 由cos2a=2cos^2a-1 而来
=1/2(cos20°+cos100°+cos140°+3)
=1/2(2cos(20°+100°)/2cos(20°-100°)/2+cos(180°-40°)+3)
=1/2(2cos60°cos40-cos40°+3)
=1/2(cos40°-cos40°+3)
=3/2
(2cos70°+根号3sin10°)/cos10°
数学函数化简题2cos10°-sin20°/cos20°=?sin40°+2cos70°/sin50°=?
求值:2sin50°+cos10°*（1+根号3tan10°）/cs35°cos40°+cos50°cos55°
化简求值2sin50°+cos10°(1+3tan10°)cos35°cos40°+cos50°cos55°
2sin50°cos50°怎么等于cos50°cos40°+sin50°sin40°
sin70°-cos70°/sin70°+cos70°=多少啊
sin50°=a cos40°=?cos50°?
求值：（2sin80°-cos70°）/cos20°
三角函数之求值问题sin²20°+cos²50°+sin20°cos50°=sin20°cos50
求cos20°cos10
=2cos40*sin40/cos10到
1.（2cos10-sin20)/cos20=______