As we know that the maximum value of A cos α + B sin α + C is C + √(A2 + B2),
And here the minimum value is C – √(a2 + B2).
(i) Given as f(x) = 12 sin x – 5 cos x
Here, A = -5, B = 12 and C = 0
–√((-5)2 + 122) ≤ 12 sin x – 5 cos x ≤ √((-5)2 + 122)
–√(25 + 144) ≤ 12 sin x – 5 cos x ≤ √(25 + 144)
–√169 ≤ 12 sin x – 5 cos x ≤ √169
-13 ≤ 12 sin x – 5 cos x ≤ 13
Thus, the maximum and minimum values of f(x) are 13 and -13 respectively.
(ii)12 cos x + 5 sin x + 4
Given as f(x) = 12 cos x + 5 sin x + 4
Here, A = 12, B = 5 and C = 4
4 – √(122 + 52) ≤ 12 cos x + 5 sin x + 4 ≤ 4 + √(122 + 52)
4 – √(144 + 25) ≤ 12 cos x + 5 sin x + 4 ≤ 4 + √(144 + 25)
4 – √169 ≤ 12 cos x + 5 sin x + 4 ≤ 4 + √169
-9 ≤ 12 cos x + 5 sin x + 4 ≤ 17
Thus, the maximum and minimum values of f(x) are -9 and 17 respectively.
(iii)5 cos x + 3 sin (π/6 – x) + 4
Given as f(x) = 5 cos x + 3 sin (π/6 – x) + 4
As we know that, sin (A – B) = sin A cos B – cos A sin B
f(x) = 5 cos x + 3 sin (π/6 – x) + 4
= 5cos x + 3 (sin π/6 cos x – cos π/6 sin x) + 4
= 5 cos x + 3/2 cos x – 3√3/2 sin x + 4
= 13/2 cos x – 3√3/2 sin x + 4
Therefore, here A = 13/2, B = – 3√3/2, C = 4
4 – √[(13/2)2 + (-3**√**3/2)2] ≤ 13/2 cos x – 3√3/2 sin x + 4 ≤ 4 + √[(13/2)2 + (-3**√**3/2)2]
4 – √[(169/4) + (27/4)] ≤ 13/2 cos x – 3√3/2 sin x + 4 ≤ 4 + √[(169/4) + (27/4)]
4 – 7 ≤ 13/2 cos x – 3√3/2 sin x + 4 ≤ 4 + 7
-3 ≤ 13/2 cos x – 3√3/2 sin x + 4 ≤ 11
Thus, the maximum and minimum values of f(x) are -3 and 11 respectively.
(iv) sin x – cos x + 1
Given as f(x) = sin x – cos x + 1
Therefore, here A = -1, B = 1 And c = 1
1 – √[(-1)2 + 12] ≤ sin x – cos x + 1 ≤ 1 + √[(-1)2 + 12]
1 – √(1 + 1) ≤ sin x – cos x + 1 ≤ 1 + √(1 + 1)
1 – √2 ≤ sin x – cos x + 1 ≤ 1 + √2
Thus, the maximum and minimum values of f(x) are 1 – √2 and 1 + √2 respectively.
