(1)求证;sinαcosα/sin²α-cos²α=-1/2tan2α(2)求sin50°(1+√3tan10°)的值
问题描述:
(1)求证;sinαcosα/sin²α-cos²α=-1/2tan2α(2)求sin50°(1+√3tan10°)的值
答
-1/2tan2α
=- tana/(1-tan^a)
=-cos^a*tana/(cos^a-sin^a)
=sinαcosα/sin²α-cos²α
右边=左边
∴sinαcosα/sin²α-cos²α=-1/2tan2α
(2)求
sin50°(1+√3tan10°)
=sin50°(cos10°+√3sin10°)/cos10°
=2sin50°(1/2 cos10°+√3/2 sin10°)/cos10°
=2sin50°sin40°/cos10°
=2cos40°sin40°/cos10°
=sin80°/cos10°
=1