sin2α=2tanα/(1+tan²α) ,cos2α=(1-tan²α)/(1+tan²α)问这两条式怎么来的
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![sin2α=2tanα/(1+tan²α) ,cos2α=(1-tan²α)/(1+tan²α)问这两条式怎么来的](/uploads/image/z/2615497-25-7.jpg?t=sin2%CE%B1%3D2tan%CE%B1%2F%281%2Btan%26sup2%3B%CE%B1%29+%2Ccos2%CE%B1%3D%281-tan%26sup2%3B%CE%B1%29%2F%281%2Btan%26sup2%3B%CE%B1%29%E9%97%AE%E8%BF%99%E4%B8%A4%E6%9D%A1%E5%BC%8F%E6%80%8E%E4%B9%88%E6%9D%A5%E7%9A%84)
sin2α=2tanα/(1+tan²α) ,cos2α=(1-tan²α)/(1+tan²α)问这两条式怎么来的
sin2α=2tanα/(1+tan²α) ,cos2α=(1-tan²α)/(1+tan²α)
问这两条式怎么来的
sin2α=2tanα/(1+tan²α) ,cos2α=(1-tan²α)/(1+tan²α)问这两条式怎么来的
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)
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
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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)
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