lgzrelv的部落格

標題:

有a.maths 唔識做

發問:

1) solve the equation sin 5θ/sinθ - cos5θ/cosθ = 2 for 0°≦θ≦360° 2) solve the equation sin^4θ + cos^4θ = sin2θfor 0°≦θ≦360

最佳解答:

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sin 5θ/sinθ - cos5θ/cosθ = 2 sin5θcosθ-cos5θsinθ=2sinθcosθ 1/2[sin6θ+sin4θ]-1/2[sin6θ-sin4θ]=2sinθcosθ 2sin6θ=2sin2θ sin6θ=sin2θ sin6θ-sin2θ=0 2cos4θcos2θ=0 cos4θ=0 or cos2θ=0 θ=22.5°, 67.5°, 112.5°, 157.5°, 202.5°, 247.5°, 292.5° or 337.5° sin^4θ + cos^4θ = sin2θ (sin2θ+cos2θ)2=sin^4θ+cos^4θ+2sin2θcos2θ ∴sin^4θ+cos^4θ=1-2sin2θcos2θ ∴1-2sin2θcos2θ=2sinθcosθ 1-2y2=2y (Let y=sinθcosθ) 2y2+2y-1=0 y=[-2±sqrt(12))]/4 y=-2±2sqrt(3)/4 sinθcosθ=(-1+sqrt3)/2 or (-1-sqrt3)/2 (rej) (1/2)sin2θ=(sqrt3-1)/2 sin2θ=sqrt3-1 2θ=47.0586°, 132.9414°, 407.0856° or 492.9414° θ=23.53° or 66.47° , 203.53° or 246.47°

其他解答:

sin^4θ + cos^4θ = sin2θ (sin^2θ)2 +( cos^2θ)2 = sin2θ (sin^2θ + cos^2θ)-2sinθcosθ = sin2θ 1-sin2θ=sin2θ 1/2=sin2θ 2θ=30° or 150° θ=15°or 75° (general: 2θ=180°n+(-1)^n 30° ) (general:θ=90°n+(-1)^n 15°)0D7DAC4E7B8CAAC5