(tan10°-根号3)*(sin80°/cos40°
问题描述:
(tan10°-根号3)*(sin80°/cos40°
答
(tan10°-√3)*(sin80°/cos40°)=(tan10°-√3)*(sin80°/cos40°)=(sin80/cos40)(cot80-√3)=(sin80/cos40)*(cos80/sin80)-(sin80/cos40)*√3 =cos80/cos40-√3sin80/cos40 =-(√3sin80-cos80)/cos40=-2(√3/2*s...为什么可以用cot来算tan10°=tan(90°-80°)=cot80°方便计算那也是tan80°啊。。。。。还是不明白公式六: π/2±α及3π/2±α与α的三角函数值之间的关系: sin(π/2+α)= cosα cos(π/2+α)= -sinα tan(π/2+α)= -cotα cot(π/2+α)= -tanα sin(π/2-α)= cosα cos(π/2-α)= sinα tan(π/2-α)= cotα cot(π/2-α)= tanα sin(3π/2+α)= -cosα cos(3π/2+α)= sinα tan(3π/2+α)= -cotα cot(3π/2+α)= -tanα sin(3π/2-α)= -cosα cos(3π/2-α)= -sinα tan(3π/2-α)= cotα cot(3π/2-α)= tanα