Question

sin2α=2tanα/(1+tan²α) ， cos2α=(1-tan²α)/(1+tan²α)

问这两条式怎么来的

Answer 1

用a代替
1/sin2a=(sin²a+cos²a)/2sinacosa
=sin²a/2sinacosa+cos²a/2sinacosa
=(1/2)(sina/cosa+cosa/sina)
=(1/2)(tana+1/tana)
=(tan²a+1)/(2tana)
sin2a=2tana/(tan²a+1)

cos2a=cos²a-sin²a
=(cos²a-sin²a)/1
=(cos²a-sin²a)/(cos²a+sin²a)
上下除cos²a
sin²a/cos²a=tan²a
所以cos2a=(1-tan²a)/(1+tan²a)

Answer 2

sin2α= 2tanα/(1+tan^2α);
cos2α=(1-tan^2α)/(1+tan^2α)

使用sin^2α+cos^2α=1

sin2α=2sinαcosα
=2sinαcosα/（sin^2α+cos^2α）,【上下同除以cos^2a】
=2tanα/(1+tan^2α);

cos2α=cos^2α-sin^2α
=(cos^2α-sin^2α)/（sin^2α+cos^2α）,【上下同除以cos^2a】
=(1-tan^2α)/(1+tan^2α)