※ 引述《Ericdion ( 由心出發 )》之銘言:
: 兩題,第296與309
: 想了幾天還是沒找到方法
: 請高手指教,感恩!!!
: http://i.imgur.com/1tTQBn5.jpg
: http://i.imgur.com/3XhdqAp.jpg
: -----
: Sent from JPTT on my Asus PadFone T004.
296. tanα+tanβ= a, tanαtanβ=b
(cosα)^2 (sinβ)^2
(cosα)^2-(sinβ)^2 = --------------------- - ---------------------
(cosα)^2+(sinα)^2 (cosβ)^2+(sinβ)^2
1 (tanβ)^2
= -------------- - ---------------
1+(tanα)^2 1+(tanβ)^2
1+(tanβ)^2 - (tanβ)^2 -(tanαtanβ)^2
= ------------------------------------------
[1+(tanα)^2][1+(tanβ)^2]
1-(tanαtanβ)^2
= --------------------------------------
1+(tanα)^2+(tanβ)^2+(tanαtanβ)^2
1-b^2
= ---------------
1+a^2-2b+b^2
--
※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 123.110.197.79
※ 文章網址: https://www.ptt.cc/bbs/Math/M.1426693857.A.21B.html