Question

不查表求值：(

1

tan5°

-tan5°)•

1 - cos20°

3 sin50° - cos50°

．

Answer 1

分析：利用正切函數(shù)化為正弦函數(shù)、余弦函數(shù)，二倍角公式，兩角差正弦函數(shù)化簡，即可得到結(jié)果．

解答：解：(

1

tan5°

-tan5°)•

1 - cos20°

3 sin50° - cos50°

=(

cos5°

sin5°

-

sin5°

cos5°

)•

1 - cos20°

3 sin50° - cos50°

=

cos10°

sin10°

•

1 - cos20°

1 2 ( 3 sin50° - cos50°)

=

cos10°

sin10°

•

1 - 1+2sin 210°

sin20°

=1．
故答案為：1．

點(diǎn)評：本題是基礎(chǔ)題，考查三角函數(shù)的化簡求值，切割化弦，二倍角公式的應(yīng)用，考查計算能力．