The maximum value of 3cosθ + 5sin(θ - π/6) for any real value of θ is : (A) √19 (B) √79/2 (C) √31 (D

Q: Wait, why can't I just find the maximum of the cosine part and the sine part separately and add them?

This is a classic trap! The maximum of $3\cos \theta$ is $3$ and the maximum of $5\sin(\theta - \pi/6)$ is $5$. If you add them to get $8$, you are assuming that both terms reach their maximum at the exact same value of $\theta$. However, $\cos \theta$ and $\sin(\theta - \pi/6)$ are out of phase. They cannot both be at their peak simultaneously. We must combine them into a single trigonometric function first.

Q: Why does the maximum value of $a \cos \theta + b \sin \theta$ equal $\sqrt{a^2 + b^2}$?

Think of this geometrically. If you multiply and divide by $\sqrt{a^2 + b^2}$, you get $\sqrt{a^2 + b^2} (\frac{a}{\sqrt{a^2 + b^2}} \cos \theta + \frac{b}{\sqrt{a^2 + b^2}} \sin \theta)$. If you let $\cos \phi = \frac{a}{\sqrt{a^2 + b^2}}$ and $\sin \phi = \frac{b}{\sqrt{a^2 + b^2}}$, the expression becomes $\sqrt{a^2 + b^2} \cos(\theta - \phi)$. Since the maximum value of the cosine function is $1$, the maximum value of the entire expression is simply $\sqrt{a^2 + b^2}$.

Q: Is there a way to solve this using calculus?

Absolutely! You could define $f(\theta) = 3\cos \theta + 5\sin(\theta - \pi/6)$, find the derivative $f'(\theta)$, set it to zero to find the critical points, and then evaluate the function at those points. While this is mathematically sound, it is significantly more time-consuming than the algebraic identity method. In JEE Advanced, time is your most precious resource, so the identity method is the preferred 'pro' approach.

Question

The maximum value of $3\cos 	heta + 5\sin(	heta - \pi/6)$ for any real value of $	heta$ is :

Options:

(A) $\sqrt{19}$

(B) $\sqrt{79}/2$

(C) $\sqrt{31}$

(D) $\sqrt{34}$

Sigma Percentile · Accepted Answer

### The Art of Harmonic Unification

Welcome, future engineers. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of trigonometric functions. We are presented with the expression $3\cos 	heta + 5\sin(	heta - \frac{\pi}{6})$.

Your intuition might scream, 'Just maximize the parts!' But in the world of JEE Advanced, intuition without rigor is a dangerous path. Let us embark on a journey to tame this expression.

### Phase 1: The Mismatched Gears

Imagine you are an engineer trying to synchronize two rotating shafts. One shaft is rotating at angle $	heta$, and the other is offset by $\frac{\pi}{6}$. You cannot simply add their outputs because they are not in phase.

This is exactly what we face here. We have a $\cos 	heta$ term and a $\sin(	heta - \frac{\pi}{6})$ term. They are speaking different languages.

To solve this, we must bring them into the same domain. We need to break down that compound angle $	heta - \frac{\pi}{6}$ using the sine difference identity:

$$\sin(A - B) = \sin A \cos B - \cos A \sin B$$

This is our master key.

### Phase 2: The Expansion

Let us apply the identity carefully. We set $A = 	heta$ and $B = \frac{\pi}{6}$. The term $5\sin(	heta - \frac{\pi}{6})$ transforms into:

$$5\left(\sin 	heta \cos\left(\frac{\pi}{6}ight) - \cos 	heta \sin\left(\frac{\pi}{6}ight)ight)$$

Now, we recall our standard trigonometric values: $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ and $\sin(\frac{\pi}{6}) = \frac{1}{2}$. Substituting these values, our expression becomes:

$$5\left(\frac{\sqrt{3}}{2} \sin 	heta - \frac{1}{2} \cos 	hetaight)$$

### Phase 3: The Algebraic Synthesis

Now, we bring back the original $3\cos 	heta$ and distribute the $5$. Our full expression is now:

$$3\cos 	heta + \frac{5\sqrt{3}}{2} \sin 	heta - \frac{5}{2} \cos 	heta$$

Notice how the chaos is beginning to organize itself? We have two terms involving $\cos 	heta$. Let us group them:

$$\left(3 - \frac{5}{2}ight) \cos 	heta + \frac{5\sqrt{3}}{2} \sin 	heta$$

Since $3 - 2.5 = 0.5$, or $\frac{1}{2}$, we arrive at the beautiful, simplified form:

$$\frac{1}{2} \cos 	heta + \frac{5\sqrt{3}}{2} \sin 	heta$$

This is the standard form $a \cos 	heta + b \sin 	heta$, where $a = \frac{1}{2}$ and $b = \frac{5\sqrt{3}}{2}$.

### Phase 4: The Geometric Elegance

Why do we love the form $a \cos 	heta + b \sin 	heta$? Because it represents the projection of a vector onto a rotating axis. Geometrically, this expression is equivalent to the hypotenuse of a right-angled triangle with legs $a$ and $b$, multiplied by a phase-shifted sine or cosine wave.

The maximum value of this expression is simply the length of that hypotenuse: $\sqrt{a^2 + b^2}$.

Let us calculate this final step. We need to find:

$$\sqrt{\left(\frac{1}{2}ight)^2 + \left(\frac{5\sqrt{3}}{2}ight)^2}$$

Squaring the terms, we get:

$$\frac{1}{4} + \frac{25 	imes 3}{4} = \frac{1}{4} + \frac{75}{4} = \frac{76}{4}$$

Simplifying this fraction, we get $19$. Thus, the maximum value is **$\sqrt{19}$**.

### Conclusion

Look at what we have achieved. We started with a confusing, mixed-angle expression and, through the power of trigonometric identities and algebraic grouping, reduced it to a simple geometric constant.

This is the essence of JEE Advanced physics and mathematics: taking complexity and finding the underlying simplicity. Never fear the messy expression; just find the identity that brings order to the chaos. You have mastered the harmonic unification. Keep practicing, and keep questioning.

Animated Solution for Mathematics - Trigonometry: The maximum value of $3 cos θ + 5 sin (θ - π /6)$ for any real value of $θ$ is :

Select Answer:

Visualized Solution

Objective: Maximize the Expression

Recall: Sine Difference Identity

Expand the Sine Term

Substitute Standard Values

Reconstruct the Expression

Distribute the Constant

Group and Simplify

Maximum Value Formula

Identify Coefficients $a$ and $b$

Square the Coefficients

Sum and Square Root

Final Answer

The Sigma Insight: Trigonometric Ratios and Identities

The Art of Harmonic Unification

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The maximum value of 3cosθ+5sin(θ−π/6) for any real value of θ is :