標題:
發問:
Solve the eqt for 0-360` 1. cos(15+x)cos(x-15)-sinxsin(x-30)=cos165 2.tan(30+x)tan(30-x)=1-2cos2x 3.sin3xsin^3 X+cos3xcos^3x =1/8 P.S所有的數字都代表度數!
最佳解答:
1. cos(15°+x)cos(x-15°)-sinxsin(x-30°)=cos165° cos(15°+x)cos(x-15°) - sinxsin(x-30°) = cos165° (1/2)[cos(15°+x+x-15°) + cos(15°+x-x+15°)] - (1/2)[cos(x-x+30°) - cos(x+x-30°)] = cos165° (1/2)[cos2x + cos30° - cos30° + cos(2x-30°)] = cos165° cos2x + cos(2x-30°) = 2cos165° 2cos[(2x+2x-30°)/2]cos[(2x-2x+30°)/2] = 2cos(180°-15°) cos(2x-15°) cos(15°) = -cos(15°) cos(2x-15°) = -1 2x-15° = 180° or 2x-15° = 540° x = 97.5° or 277.5° ============================== 2.tan(30°+x)tan(30°-x) = 1 - 2cos2x tan(30°+x)tan(30°-x) = 1 - 2cos2x sin(30°+x)sin(30°-x) / cos(30°+x)cos(30°-x) = 1 - 2cos2x (1/2)[cos(30°+x-30°+x) - cos(30°+x+30°-x)] / (1/2)[cos(30°+x+30°-x) + cos(30°+x-30°+x)] = 1 - 2cos2x (cos2x - cos60°) / (cos60° + cos2x) = 1 - 2cos2x (cos2x - 1/2) / (1/2 + cos2x) = 1 - 2cos2x (2cos2x - 1) / (1 + 2cos2x) = 1 - 2cos2x 【Assume 2cos2x + 1 ≠ 0】 -(1 - 2cos2x) = (1 - 2cos2x)(1 + 2cos2x) (1 - 2cos2x)(1 + 2cos2x + 1) = 0 (1 - 2cos2x)(2cos2x + 2) = 0 1 - 2cos2x = 0 or 2cos2x + 2 = 0 cos2x = 1/2 or cos2x = -1 For cos2x = 1/2 2x = 60° or 2x = 300° or 2x = 420° or 2x = 660° x = 30° or x = 150° or x = 210° or x = 330° For cos2x = -1 2x = 180° or 2x = 540° x = 90° or x = 270° So the solution is x = 30° or 90° or 150° or 210° or 270° or 330° ============================== 3.sin3xsin3X + cos3xcos3x = 1/8 sin3x sin3x + cos3xcos3x = 1/8 sin3x sinx(sin2x) + cos3x cosx (cos2x) = 1/8 (1/2)[cos(3x-x) - cos(3x+x)] sin2x + (1/2)[cos(3x+x) + cos(3x-x)] cos2x = 1/8 (cos2x - cos4x)sin2x + (cos4x + cos2x)cos2x = 1/4 cos2x(sin2x + cos2x) + cos4x(cos2x - sin2x) = 1/4 cos2x(1) + cos4x(cos2x) = 1/4 cos2x[1 + cos4x] = 1/4 cos2x[1 + 2cos22x - 1] = 1/4 2cos32x = 1/4 cos32x = 1/8 cos2x = 1/2 2x = 60° or 2x = 300° or 2x = 420° or 2x = 660° x = 30° or x = 150° or x = 210° or x = 330° ============================== 2007-02-07 23:01:38 補充： 小小補充：應用到的公式有sin2x + cos2x = 1sin2x = 2sinx cosxcos2x = 2cos2x - 1 = cos2x - sin2xcosAcosB = (1/2)[cos(A+B) + cos(A-B)]sinAsinB = (1/2)[cos(A-B) - cos(A+B)]
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a.maths Q.發問:
Solve the eqt for 0-360` 1. cos(15+x)cos(x-15)-sinxsin(x-30)=cos165 2.tan(30+x)tan(30-x)=1-2cos2x 3.sin3xsin^3 X+cos3xcos^3x =1/8 P.S所有的數字都代表度數!
最佳解答:
1. cos(15°+x)cos(x-15°)-sinxsin(x-30°)=cos165° cos(15°+x)cos(x-15°) - sinxsin(x-30°) = cos165° (1/2)[cos(15°+x+x-15°) + cos(15°+x-x+15°)] - (1/2)[cos(x-x+30°) - cos(x+x-30°)] = cos165° (1/2)[cos2x + cos30° - cos30° + cos(2x-30°)] = cos165° cos2x + cos(2x-30°) = 2cos165° 2cos[(2x+2x-30°)/2]cos[(2x-2x+30°)/2] = 2cos(180°-15°) cos(2x-15°) cos(15°) = -cos(15°) cos(2x-15°) = -1 2x-15° = 180° or 2x-15° = 540° x = 97.5° or 277.5° ============================== 2.tan(30°+x)tan(30°-x) = 1 - 2cos2x tan(30°+x)tan(30°-x) = 1 - 2cos2x sin(30°+x)sin(30°-x) / cos(30°+x)cos(30°-x) = 1 - 2cos2x (1/2)[cos(30°+x-30°+x) - cos(30°+x+30°-x)] / (1/2)[cos(30°+x+30°-x) + cos(30°+x-30°+x)] = 1 - 2cos2x (cos2x - cos60°) / (cos60° + cos2x) = 1 - 2cos2x (cos2x - 1/2) / (1/2 + cos2x) = 1 - 2cos2x (2cos2x - 1) / (1 + 2cos2x) = 1 - 2cos2x 【Assume 2cos2x + 1 ≠ 0】 -(1 - 2cos2x) = (1 - 2cos2x)(1 + 2cos2x) (1 - 2cos2x)(1 + 2cos2x + 1) = 0 (1 - 2cos2x)(2cos2x + 2) = 0 1 - 2cos2x = 0 or 2cos2x + 2 = 0 cos2x = 1/2 or cos2x = -1 For cos2x = 1/2 2x = 60° or 2x = 300° or 2x = 420° or 2x = 660° x = 30° or x = 150° or x = 210° or x = 330° For cos2x = -1 2x = 180° or 2x = 540° x = 90° or x = 270° So the solution is x = 30° or 90° or 150° or 210° or 270° or 330° ============================== 3.sin3xsin3X + cos3xcos3x = 1/8 sin3x sin3x + cos3xcos3x = 1/8 sin3x sinx(sin2x) + cos3x cosx (cos2x) = 1/8 (1/2)[cos(3x-x) - cos(3x+x)] sin2x + (1/2)[cos(3x+x) + cos(3x-x)] cos2x = 1/8 (cos2x - cos4x)sin2x + (cos4x + cos2x)cos2x = 1/4 cos2x(sin2x + cos2x) + cos4x(cos2x - sin2x) = 1/4 cos2x(1) + cos4x(cos2x) = 1/4 cos2x[1 + cos4x] = 1/4 cos2x[1 + 2cos22x - 1] = 1/4 2cos32x = 1/4 cos32x = 1/8 cos2x = 1/2 2x = 60° or 2x = 300° or 2x = 420° or 2x = 660° x = 30° or x = 150° or x = 210° or x = 330° ============================== 2007-02-07 23:01:38 補充： 小小補充：應用到的公式有sin2x + cos2x = 1sin2x = 2sinx cosxcos2x = 2cos2x - 1 = cos2x - sin2xcosAcosB = (1/2)[cos(A+B) + cos(A-B)]sinAsinB = (1/2)[cos(A-B) - cos(A+B)]
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