COS(-100°)=a,则tan80°=?

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COS(-100°)=a,则tan80°=?COS(-100°)=a,则tan80°=?COS(-100°)=a,则tan80°=?cos(-100°)=cos100°=-cos80°=a,则cos8

COS(-100°)=a,则tan80°=?
COS(-100°)=a,则tan80°=?

COS(-100°)=a,则tan80°=?
cos(-100°)=cos100°=-cos80°=a,则cos80°=-a,因sin²80°=1-cos²80°=1+a²且sin80°>0,则sin80°=√(1-a²).tan80°=sin80°/cos80°=[√(1-a²)]/(-a)

cos(-100º)=cos100º=cos(180º-80º)=-cos80º=a
cos80º=-a
tan80º=sin80º/cos80º=[√(1-cos²80º)]/cos80º=[√(1-a²)]/(-a)=-[√(1-a²)]/a

COS(-100°)=-cos80°=a
cos80°=-a
sin80°=√[1-(-a0^2]=√(1-a^2)
tan80°=sin80°/cos80°
=√(1-a^2)/(-a )
=-√(1-a^2)/a

cos100º=a cos80º=-a tan80º=√(1-cos²80º)/ºcos80º=-√(1-a²)/²a

tan80°= tan (180° -100°) = - tan100° = - sin100°/ cos100°
cos100°= cos(-100°)=a
sin100° = (1-a^2)^2
tan80°=(1-a^2)^2 /a
因为 COS(-100°)=a,所以 1< a < 0, tan80°= - (1-a^2)^2/a