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Question: Maximum value of 3 cos θ+ 4 sin $(θ-\frac{π}{6})$...

Maximum value of 3 cos θ+ 4 sin (θ−π6)(θ-\frac{π}{6})

A

762\frac{\sqrt{76}}{2}

B

132\frac{\sqrt{13}}{2}

C

13\sqrt{13}

D

None of these

Answer

13\sqrt{13}

Explanation

Solution

Let the given expression be EE.

E=3cos⁡θ+4sin⁡(θ−π6)E = 3 \cos \theta+ 4 \sin (\theta-\frac{\pi}{6})

First, expand sin⁡(θ−π6)\sin (\theta-\frac{\pi}{6}) using the identity sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin(A-B) = \sin A \cos B - \cos A \sin B:

sin⁡(θ−π6)=sin⁡θcos⁡π6−cos⁡θsin⁡π6\sin (\theta-\frac{\pi}{6}) = \sin \theta \cos \frac{\pi}{6} - \cos \theta \sin \frac{\pi}{6}

We know cos⁡π6=32\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} and sin⁡π6=12\sin \frac{\pi}{6} = \frac{1}{2}.

So, sin⁡(θ−π6)=sin⁡θ⋅32−cos⁡θ⋅12\sin (\theta-\frac{\pi}{6}) = \sin \theta \cdot \frac{\sqrt{3}}{2} - \cos \theta \cdot \frac{1}{2}

Substitute this back into the expression for EE:

E=3cos⁡θ+4(32sin⁡θ−12cos⁡θ)E = 3 \cos \theta + 4 \left( \frac{\sqrt{3}}{2} \sin \theta - \frac{1}{2} \cos \theta \right)

E=3cos⁡θ+23sin⁡θ−2cos⁡θE = 3 \cos \theta + 2\sqrt{3} \sin \theta - 2 \cos \theta

Combine the cos⁡θ\cos \theta terms:

E=(3−2)cos⁡θ+23sin⁡θE = (3-2) \cos \theta + 2\sqrt{3} \sin \theta

E=cos⁡θ+23sin⁡θE = \cos \theta + 2\sqrt{3} \sin \theta

The maximum value of an expression of the form acos⁡x+bsin⁡xa \cos x + b \sin x is a2+b2\sqrt{a^2 + b^2}.

In this case, a=1a=1 and b=23b=2\sqrt{3}.

Maximum value of E=12+(23)2E = \sqrt{1^2 + (2\sqrt{3})^2}

Maximum value of E=1+(4×3)E = \sqrt{1 + (4 \times 3)}

Maximum value of E=1+12E = \sqrt{1 + 12}

Maximum value of E=13E = \sqrt{13}