有a.maths 唔識做

2008-06-01 2:23 am
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

回答 (2)

2008-06-01 8:19 am
✔ 最佳答案
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°
2008-06-01 2:37 am
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°)


收錄日期: 2021-04-23 20:30:55
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