a) If 7sin X - 24cos X is expressed in the form r sin(X - A) where r > 0
and 0 < A < 90, find r and A.
b) Let y =14sin X - 48sin X + 14,
Using the result in a) , find the maximum and minimum values of y.
Find also the general values of X at which y attains its maximum.(0<=X<=180)
c) P and Q are the two acute angles satisfying the equation
6cos^2_X - 5cos X + 1=0
Without solving the equation, show that cos(P+Q)/2 + cos (P-Q)/2 = root of 2
[Hint: cos^_x/2 = 1/2(1+cos x)
三角學問題HKCEE1987
2013-03-31 6:35 pm
回答 (1)
2013-04-03 10:31 pm
✔ 最佳答案
y =14sin X - 48sin X + 14 <-- "14sin X - 48sin X" 有沒有打錯??
2013-04-03 14:31:22 補充：
(a) 7sin X - 24cos X = r sin(X - A) 7sin X - 24cos X = r (sinXcosA – sinAcosX) 7sin X - 24cos X = (r cosA)sinX – (r sinA)cosX 7 = r cosA --- (i) 24 = r sinA --- (ii) (ii)÷(i), tanA = 24/7 -- 0º < A < 90º A = 73.7º (1.29 rad) (Correct to 3 sig. fig) (r cosA)²+ (r sinA)²= 7² + 24² r²(sin²A + cos²A) = 625 r = 25 (b) y = 14sin X - 48cos X + 14 y = 2(7sinX - 24cosX) + 14 y = 50sin(X-A) + 14 The max. value of y — the max. value of sinθ is always 1, no matter what θis = 50(1) + 14 = 64 The min. value of y = 50(-1) + 14 = -36 When y attains its maximum, 64 = 50sin(X-73.7º) + 14 sin(X-73.7º) = 1 X = 163.7º (2.86 rad) (Correct to 3 sig. fig)¹ Of course you can calculate dy/dx: dy/dx = 50cos(X-A) When dy/dx = 0, cos(X-A) = 0 X-A = 90º or 270º X = 163.7º or 343.7º(rejected) (c) LHS = cos(P+Q)/2 + cos (P-Q)/2 = 2cos(P/2)cos(Q/2) = √[4cos²(P/2)cos²(Q/2)] = √[(1+cosP)(1+cosQ)] = √[(cosPcosQ) + (cosP + cosQ) + 1] = √(1/6+5/6+1) —Sum of root and Product of root = √2 = RHS
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