Q.Prove that $f(x) = \sin x + \sqrt{3}\,\cos x$ has maximum value at $x = \dfrac{\pi}{6}$.

Concept: Maximum Value of Sine-Cosine Expression

Any expression of the form $a\sin x + b\cos x$ can be rewritten as $R\sin(x+\phi)$, where $R = \sqrt{a^2+b^2}$ and $\phi = \tan^{-1}(b/a)$. The maximum value is $R$, attained when $\sin(x+\phi)=1$.

Steps:

1. Here $a=1$, $b=\sqrt{3}$. Compute $R = \sqrt{1^2+(\sqrt{3})^2} = \sqrt{1+3}=2$.
2. Write $f(x) = 2\left(\frac{1}{2}\sin x + \frac{\sqrt{3}}{2}\cos x\right) = 2\sin\left(x+\frac{\pi}{3}\right)$. …

The key insight here is that any expression of the form $a\sin x + b\cos x$ can be compressed into a single trigonometric function. This isn't just a trick — it reflects the fact that sine and cosine are just phase-shifted versions of each other. Adding them with different coefficients produces another sine wave with a different amplitude and phase.

For $f(x) = \sin x + \sqrt{3}\cos x$, we have $a = 1$ and $b = \sqrt{3}$. The amplitude of the combined wave is $R = \sqrt{a^2 + b^2} = \sqrt{1 + 3} = 2$. So the maximum possible value of $f(x)$ is $2$, and we need to find the $x$ where this peak occurs.

1. Rewrite in the form $R\sin(x + \phi)$. We want: $\sin x + \sqrt{3}\cos x = R\sin(x + \phi)$. Using the sine addition formula:

Matching coefficients with $1\cdot\sin x + \sqrt{3}\cdot\cos x$ gives:

2. Find $R$ and $\phi$. Squaring and adding: $R^2(\cos^2\phi + \sin^2\phi) = 1^2 + (\sqrt{3})^2 = 4$, so $R = 2$ (positive amplitude). Then $\cos\phi = \frac{1}{2}$ and $\sin\phi = \frac{\sqrt{3}}{2}$. The angle $\phi$ that satisfies both is $\phi = \frac{\pi}{3}$ (since $\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$ and $\cos\frac{\pi}{3} = \frac{1}{2}$). Therefore:

3. Find the maximum. The sine function reaches its maximum value of $1$ when its argument equals $\frac{\pi}{2} + 2n\pi$ (for integer $n$). So:

Solving for $x$:

The smallest positive $x$ where this occurs is $x = \frac{\pi}{6}$. …

Method: Maximizing $a\sin x + b\cos x$ using the Auxiliary Angle Technique

Any linear combination of sine and cosine can be collapsed into a single sine wave with a known amplitude — this turns a calculus optimization problem into simple trigonometric reasoning.

Steps

Step 1: Identify the coefficients and compute the amplitude

For an expression $a\sin x + b\cos x$, compute

This $R$ is the maximum possible value the expression can ever take, before you even find where.

Step 2: Find the auxiliary angle $\phi$

Choose $\phi$ so that $\cos\phi = \dfrac{a}{R}$ and $\sin\phi = \dfrac{b}{R}$ (equivalently $\tan\phi = b/a$, picking the correct quadrant from the signs of $a$ and $b$). Then, using the sine addition formula in reverse,

Step 3: Use the known maximum of sine

The sine function reaches its maximum value of $1$ exactly when its argument equals $\dfrac{\pi}{2} + 2n\pi$. Set

$x + \phi = \frac{\pi}{2} + 2n\pi$ …

Common Mistakes

Mistake 1: Setting $f'(x)=0$ and stopping without checking it's a maximum, not a minimum

Differentiating directly gives $f'(x)=\cos x-\sqrt3\sin x$, and setting this to zero leads to $\tan x=\tfrac{1}{\sqrt3}$, i.e. $x=\pi/6$ — but $\tan x=1/\sqrt3$ also has other solutions (e.g. $x=\pi/6+\pi$) where $f$ is actually minimised. Why it's wrong: a zero derivative only marks a critical point; without a second-derivative or sign check, the student can't tell a peak from a valley. Correct approach: either verify $f''(\pi/6)<0$, or use the $R\sin(x+\phi)$ rewrite, where the maximum of a sine function is unambiguous — it occurs exactly when the argument is $\pi/2+2n\pi$.

Mistake 2: Getting the amplitude-phase formula backwards …