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a.maths question

發問:

Please solve this equation for 0degree<=x<=360degree sin3x+cos3x=sin2x+cos2x

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最佳解答:

Using formula sin A - sin B = 2cos[(A + B)/2]sin[(A - B)/2] cosA – cosB = - 2sin[(A + B)/2]sin[(A - B)/2] cos(A+B) = cosAcosB - sinAsinB SOLUTION sin3x+cos3x=sin2x+cos2x sin3x – sin2x + cos3x – cos2x = 0 2cos[(3x + 2x)/2]sin[(3x – 2x)/2] + 2sin[(3x + 2x)/2]sin[(3x – 2x)/2] = 0 2cos(2.5x)sin(0.5x) - 2sin(2.5x)sin(0.5x) = 0 2[cos(2.5x) – sin(2.5x)]sin(0.5x) = 0 [divided by cos(2.5x)] 2[1 – tan(2.5x)]sin(0.5x) = 0 sin 0.5x = 0 or tan(2.5x) = 1 when sin(0.5x) = 0 0.5x = 0 or 0.5x = 180 So x = 0 or or = 360 when tan(2.5x) = 1 2.5x = 45 or 2.5x = 225 or 2.5x = 405 or 2.5x = 585 or 2.5x = 765 x = 18 or x = 90 or x = 162 or x = 234 or x = 306

其他解答:

Using formula sin A - sin B = 2cos[(A + B)/2]sin[(A - B)/2] cosA – cosB = - 2sin[(A + B)/2]sin[(A - B)/2] cos(A+B) = cosAcosB - sinAsinB SOLUTION sin3x+cos3x=sin2x+cos2x sin3x – sin2x + cos3x – cos2x = 0 2cos[(3x + 2x)/2]sin[(3x – 2x)/2] + 2sin[(3x + 2x)/2]sin[(3x – 2x)/2] = 0 2cos(2.5x)sin(0.5x) - 2sin(2.5x)sin(0.5x) = 0 2[cos(2.5x) – sin(2.5x)]sin(0.5x) = 0 [divided by cos(2.5x)] 2[1 – tan(2.5x)]sin(0.5x) = 0 sin 0.5x = 0 or tan(2.5x) = 1 when sin(0.5x) = 0 0.5x = 0 or 0.5x = 180 So x = 0 or or = 360 when tan(2.5x) = 1 2.5x = 45 or 2.5x = 225 or 2.5x = 405 or 2.5x = 585 or 2.5x = 765 x = 18 or x = 90 or x = 162 or x = 234 or x = 30663D0B758E2D502CC
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