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JEE Main 2019 · 12 Jan · Shift I

JEE Main 2019 Mathematics question on Trigonometry

The maximum value of
3cos⁡θ+5sin⁡(θ−π6)3\cos\theta + 5\sin\left(\theta - \frac{\pi}{6}\right)
for any real value of θ\theta is
  1. A.
    792\dfrac{\sqrt{79}}{2}
  2. B.
    34\sqrt{34}
  3. C.
    31\sqrt{31}
  4. D.
    19\sqrt{19}
Show answer and solution

Correct option: D

Solution

Expand 5sin(θ - π/6) = 5sinθcos(π/6) - 5cosθsin(π/6) = (5√3/2)sinθ - (5/2)cosθ.

So the expression becomes (3 - 5/2)cosθ + (5√3/2)sinθ = (1/2)cosθ + (5√3/2)sinθ.

The maximum value of Acosθ + Bsinθ is √(A² + B²) = √((1/2)² + (5√3/2)²) = √(1/4 + 75/4) = √(76/4) = √19.

[JEE Main 2019 · 12 Jan · Shift I]

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