RI 1998 EOY Q16
(a) 4 cos x - 3 sin x in the form R cos(x + α) where R is positive and α is acute. Hence find the maximum and minimum values of (4 cos x - 3 sin x)² and state the values of x between 0° and 360° when these values occur. [8]
(b) Given that tan A = -3/4 and sin 2B = -8/17 where A and 2B are in the same quadrant and both between 0° and 360°, find, without using tables or calculator, the value of
(i) sin (A - 2B)
(ii) cos B
[4]
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Answer:
(a) 4 cos x - 3 sin x = R cos(x + α)
4 cos x - 3 sin x = R cos x cos α - R sin x sin α
By comparing coefficients,
R cos α = 4 ---- (1)
R sin α = 3 ---- (2)
(1)² + (2)²: (R cos α)² + (R sin α)² = 4² + 3²
R² cos² α + R² sin² α = 16 + 9
R² (cos² α + sin² α) = 25
R² = 25
R = 5
(2)/(1): R sin α / R cos α = 3 / 4
tan α = 3 / 4
α ≈ 36.9°
Hence, 4 cos x - 3 sin x = 5 cos (x + 36.9°)
-1 ≤ cos (x + 36.9°) ≤ 1
-5 ≤ 5 cos (x + 36.9°) ≤ 5
0 ≤ [5 cos (x + 36.9°)]² ≤ 25
0 ≤ [4 cos x - 3 sin x]² ≤ 25
Maximum value of [4 cos x - 3 sin x]² = 25
This occurs when cos (x + 36.9°) = 1 or cos (x + 36.9°) = -1 (since we square it)
so, x + 36.9° = 360° or 180° (0° not included because 0° ≤ x ≤ 360°)
x = 323.1° or 143.1°
Minimum value of [4 cos x - 3 sin x]² = 0
This occurs when cos (x + 36.9°) = 0
so, x + 36.9° = 90°, 270°
x = 53.1°, 233.1°
(b)
tan A = -3 /4
sin A = -3 /5
cos A = 4 /5
sin 2B = -8 /17
cos 2B = 15 /17
tan 2B = -8 / 15
(i) sin (A - 2B) = sin A cos 2B - cos A sin 2B
= (-3/5)(15/17) - (4/5)(-8/17)
= -9/17 + 32/85
= -13/85
(ii) cos 2B = 2 cos² B - 1
2 cos² B = 1 + 15/17
cos² B = 16/17
cos B = √(16/17) or -√(16/17)
