tanx的倍角半角公式

Question

如题所述

Answer 1

二倍角公式：
S I n ( 2 α ) = 2 s I n α · c o s α = 2 / ( t a n α + c o t α )
C o s ( 2 α ) = c o s ^ 2 ( α ) – s I n ^ 2 ( α ) = 2 c o s ^ 2 ( α ) – 1 = 1 – 2 s I n ^ 2 ( α )
T a n ( 2 α ) = 2 t a n α / [ 1 – t a n ^ 2 ( α ) ]
半角公式：1.正弦 s I n ( A / 2 ) = √ ( ( 1 – c o s A ) / 2 ) s I n ( A / 2 ) = - √ ( ( 1 -c o s A ) / 2 )
2.余弦 c o s ( A / 2 ) = √ ( ( 1 + c o s A ) / 2 ) c o s ( A / 2 ) = - √ ( ( 1 + c o s A ) / 2 )
3.正切 t a n ( A / 2 ) = √ ( ( 1 – c o s A ) / ( ( 1 + c o s A ) ) t a n ( A / 2 ) = - √ ( ( 1 -c o s A ) / ( ( 1 + c o s A )
万能公式 s I n α = 2 t a n ( α / 2 ) / [ 1 + t a n ^ ( α / 2 ) ] c o s α = [ 1 -t a n ^ ( α / 2 ) ] / 1 + t a n ^ ( α / 2 ) ] t a n α = 2 t a n ( α / 2 ) / [ 1 – t a n ^ ( α / 2 ) ]