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發問:

help..............步驟 求通解... tan(30°+x) tan(30°-x) = 1-2cos2x

最佳解答:

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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