(tan10°-根号3)*(sin80°/cos40°

问题描述:

(tan10°-根号3)*(sin80°/cos40°

(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