恆等式,急

2007-06-26 1:45 am
tanA/(1+tan^2A)+sinAcosA

回答 (3)

2007-06-26 2:10 am
✔ 最佳答案
  tanA
= -------------- + sinAcosA
  1+tan²A

   tanA
= -------------- + sinAcosA
   sin²A
 1+----------
   cos²A


    tanA
= ------------------------ + sinAcosA
  cos²A  sin²A
 -----------+----------
  cos²A   cos²A



    tanA
= ------------------------ + sinAcosA
  cos²A+sin²A
 ---------------------
   cos²A


    tanA
= ------------------------ + sinAcosA
     1
 ---------------------
   cos²A

= tanAcos²A+ sinAcosA

  sinA
= ----------xcos²A+ sinAcosA
  cosA


=sinAcosA+sinAcosA

=2sinAcosA
2007-07-07 10:17 am
解題就識, 2sinAcosA = sin2A 就唔識, 果然係高手, 唔比個負面你都唔得啦!
2007-06-26 2:00 am
tanA / (1 + tan^2 A) + sinA cosA
= tanA / (cos^2A / cos^2A + sin^2A / cos^2A) + sinA cosA
= sinA / cosA / (1 / cos^2A) + sinA cosA
= sinA / cosA x cos^2A + sinA cosA
= sinA cosA + sinA cos A
= 2 sinA cosA


收錄日期: 2021-05-02 13:47:32
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