Q.What is the maximum value of the function $\sin x+\cos x$?

The key idea is that $a\sin x + b\cos x$ can be rewritten as $R\sin(x+\phi)$, where $R = \sqrt{a^2+b^2}$ is the maximum value.

For $\sin x + \cos x$, we have $a=1$, $b=1$, so $R = \sqrt{1^2+1^2} = \sqrt{2}$. …

The key insight is that $\sin x + \cos x$ is not just a random sum — it’s the sum of two waves of the same frequency, which always produces another wave of that same frequency. The only thing that changes is the amplitude and the phase shift. So instead of guessing values of $x$, we can rewrite the expression in the form $R \sin(x + \phi)$ or $R \cos(x - \phi)$, where $R$ is the maximum possible value.

Why does this work? Any linear combination $a \sin x + b \cos x$ can be compressed into a single sine or cosine. The amplitude $R$ is $\sqrt{a^2 + b^2}$, and the phase shift $\phi$ satisfies $\tan \phi = \frac{b}{a}$ (or similar, depending on the form). Since $\sin$ and $\cos$ each range between $-1$ and $1$, their sum can go as high as $2$ only if both are $1$ at the same $x$ — but that never happens because $\sin x = 1$ when $x = \frac{\pi}{2}$, where $\cos x = 0$. So the true maximum is less than $2$, and the amplitude formula gives it exactly.

Let’s work through it step by step.

1. Set up the transformation. We want to write $\sin x + \cos x$ as $R \sin(x + \phi)$. Using the sine addition formula:

For this to equal $\sin x + \cos x$, we need:

2. Find $R$. Square both equations and add:

Since $\cos^2 \phi + \sin^2 \phi = 1$, we get $R^2 = 2$, so $R = \sqrt{2}$ (we take the positive root because amplitude is positive).

3. Find the phase shift $\phi$ (optional but helpful). From $R \cos \phi = 1$ and $R \sin \phi = 1$, we have $\cos \phi = \frac{1}{\sqrt{2}}$ and $\sin \phi = \frac{1}{\sqrt{2}}$. This means $\phi = \frac{\pi}{4}$ (or $45^\circ$). So: $\sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right).$ …

Method: Maximum of $a\sin x + b\cos x$ via the Auxiliary-Angle (R-form) Technique

Whenever a question asks for the maximum (or minimum) of a linear combination of sine and cosine of the same angle, this rewriting technique is faster and more reliable than a full derivative-based search.

Steps

Step 1: Recognise the form $a\sin x + b\cos x$ and set up the target identity.

The goal is to rewrite the expression as a single wave, $R\sin(x+\phi)$, by matching coefficients using the sine addition formula:

Step 2: Match coefficients to get two equations, then solve for the amplitude $R$.

Matching gives $R\cos\phi = a$ and $R\sin\phi = b$. Squaring and adding both equations eliminates $\phi$ using $\sin^2\phi+\cos^2\phi=1$:

Step 3: Use the bounded range of sine to read off the maximum directly. …

Common Mistakes

Mistake 1: Adding the individual maximums of $\sin x$ and $\cos x$ to get $2$.

Why it's wrong: $\sin x$ reaches its own maximum of $1$ only at $x=\frac{\pi}{2}$, where $\cos x=0$, not $1$ — the two functions never hit their individual peaks at the same $x$, so $1+1=2$ overstates the true combined maximum. Correct approach: use the amplitude formula $R=\sqrt{a^2+b^2}$, which correctly accounts for the fact the two peaks are out of phase, giving $\sqrt{2}$ here, not $2$.

Mistake 2: Forgetting the amplitude formula only applies when both terms share the same angle.

Why it's wrong: the $R=\sqrt{a^2+b^2}$ shortcut is valid for $a\sin x+b\cos x$, but it cannot be applied directly to something like $\sin x+\cos 2x$, where the arguments differ — treating it the same way would give a wrong bound. Correct approach: check that both trigonometric terms have identical arguments before applying the auxiliary-angle formula.

Mistake 3: Mixing up which of $a$ and $b$ goes with $\cos\phi$ versus $\sin\phi$ when also finding the phase angle. …