急! F4 Basic Trigonometry ex5

2012-09-29 1:29 am
請詳細步驟教我計以下幾條 :
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回答 (1)

2012-09-29 2:13 am
✔ 最佳答案
25) a) max(3 cos x – 2) = 3 max(cos x) – 2 = 3 – 2 = 1min(3 cos x – 2) = 3 min(cos x) – 2 = - 3 – 2 = - 5b) max(3 sin 2x + 4) = 3 max(sin 2x) + 4 = 3 + 4 = 7min(3 sin 2x + 4) = 3 min(sin 2x) + 4 = - 3 + 4 = 1c) max(3 sin²x + 4) = 3 max(sin²x) + 4 = 3 + 4 = 7min(3 sin²x + 4) = 3 min(sin²x) + 4 = 0 + 4 = 4d) max(3/(6 – 4 cos²x)) = 3/min(6 – 4 cos²x)= 3/(6 – 4 max(cos²x)) = 3/(6 – 4) = 3/2min(3/(6 – 4 cos²x)) = 3/max(6 – 4 cos²x)= 3/(6 – 4 min(cos²x)) = 3/(6 – 0) = 1/2 29) cos(α + β)= cos(360° - γ - δ)= cos(- γ– δ)= cos(γ + δ)

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參考: My Maths World


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