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2007-04-15 10:40 pm
help..............步驟

求通解...

tan(30°+x) tan(30°-x) = 1-2cos2x

回答 (1)

2007-04-16 1:35 am
✔ 最佳答案
             tan (30°+x) tan (30°-x) = 1-2cos2x

           sin (30°+x)  sin (30°-x)
          -----.------ = 1 - 2 cos 2x
           cos (30°+x)   cos (30°-x)

             sin (30°+x) sin (30°-x)
            ---------- = 1 - 2 cos 2x
             cos (30°+x) cos (30°-x)

(-1/2) {cos [(30°+x)+(30°-x)] - cos [(30°+x)-(30°-x)] }
---------------------- = 1 - 2 cos 2x [註1]
(1/2) {cos [(30°+x)+(30°-x)] + cos [(30°+x)-(30°-x)]

            (-1/2) (cos 60° - cos 2x)
           ----------- = 1 - 2 cos 2x
            (1/2) (cos 60° + cos 2x)

              (-1/2) (1/2 - cos 2x)
             --------- = 1 - 2 cos 2x
              (1/2) (1/2 + cos 2x)

                - (1 - 2 cos 2x)
               ------- = 1 - 2 cos 2x  [註2]
                1 + 2cos 2x

                - ( 1 - 2 cos 2x) = (1 - 2 cos 2x)(1 + 2 cos 2x)
                     0 = ( 1 - 2 cos 2x)(1 + 2 cos 2x + 1)
         ( 1 - 2 cos 2x)(1 + 2 cos 2x + 1) = 0
         1 - 2 cos 2x = 0 或 2 + 2 cos 2x = 0
           cos 2x = 1/2 或 cos 2x = -1
         2x = 360°n±60° 或 2x = 360°n±180°
          x = 180°n±30° 或 x=180°n±90° // (n=1,2,3,...)

註1 : 用積化和差公式(product-to-sum)
sin A sin B = (-1/2) [cos(A+B) - cos(A-B)]
cos A cos B = (1/2) [cos(A+B) + cos(A-B)]

註2 : 分子和分母都x4


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