The maximum distance of a point on the graph of the function $y = \sqrt{3}\,\sin x + \cos x$ from $x$-axis is ______.

The maximum distance of a point on the graph of the function y = √(3)sin x + cos x from x-axis is ___.

NCERT Exemplar · Q67

y = \sqrt{3}\,\sin x + \cos x

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Concept understanding — Maximum Value Sine Cosine

Maximum Value of Sine and Cosine – The Core Idea

Imagine a point moving around a unit circle centred at the origin. Its coordinates are $(\cos\theta, \sin\theta)$ , where $\theta$ is measured from the positive x-axis.

The farthest right the point reaches is $(1,0)$ — $\cos\theta = 1$ ; the farthest left is $(-1,0)$ — $\cos\theta = -1$ . The highest is $(0,1)$ — $\sin\theta = 1$ ; the lowest is $(0,-1)$ — $\sin\theta = -1$ . So sine and cosine never exceed $1$ or fall below $-1$ : they are bounded by the unit circle.

Important

For any real angle $\theta$ ,

$-1 \le \sin \theta \le 1 \quad \text{and} \quad -1 \le \cos \theta \le 1$

The Precise Statement

Maximum value: $1$ ; minimum value: $-1$ . Both are achieved at specific angles.

For sine:

$\sin \theta = 1$ when $\theta = 90^\circ + 360^\circ n$ (i.e. $\frac{\pi}{2} + 2\pi n$ )
$\sin \theta = -1$ when $\theta = 270^\circ + 360^\circ n$ (i.e. $\frac{3\pi}{2} + 2\pi n$ )

For cosine:

$\cos \theta = 1$ when $\theta = 0^\circ + 360^\circ n$ (i.e. $2\pi n$ )
$\cos \theta = -1$ when $\theta = 180^\circ + 360^\circ n$ (i.e. $\pi + 2\pi n$ )

Here $n$ is any integer — the pattern repeats every full rotation.

Why This Matters in Exams

Many problems ask for the maximum or minimum of expressions like $3\sin x + 4\cos x$ or $2 - 5\sin x$ . Since sine and cosine are individually trapped between $-1$ and $1$ , you can bound any linear combination.

Tip

For $a\sin\theta + b\cos\theta$ , the maximum is $\sqrt{a^2 + b^2}$ and the minimum is $-\sqrt{a^2 + b^2}$ . Derive it by rewriting as $R\sin(\theta + \phi)$ .

Common Mistake to Avoid …

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Rewrite $\sqrt{3}\sin x + \cos x$ as $R\sin(x + \phi)$ to find its amplitude; the maximum distance from the $x$ -axis is the maximum value of $|y|$ , which is 2.

The question asks for the maximum distance from the $x$ -axis, which means we need the maximum value of $|y|$ . Since the function is a combination of sine and cosine, it will oscillate above and below the $x$ -axis. The key insight is to express this combination as a single sinusoidal function whose amplitude gives us exactly what we need.

Why combine sine and cosine?

Any expression of the form $a\sin x + b\cos x$ can be written as $R\sin(x + \phi)$ where $R = \sqrt{a^2 + b^2}$ . This $R$ is the amplitude—the maximum value the function reaches. Since sine oscillates between $-1$ and $1$ , the function $R\sin(x + \phi)$ oscillates between $-R$ and $R$ , making $R$ the maximum distance from the $x$ -axis.

a\sin x + b\cos x = R\sin(x + \phi) \quad \text{where} \quad R = \sqrt{a^2 + b^2}

Solution

Identify the coefficients

In our function $y = \sqrt{3}\sin x + \cos x$ , we have $a = \sqrt{3}$ and $b = 1$ .
Calculate the amplitude

Using the formula above:

R = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2

Rewrite the function We can express the function as $y = 2\sin(x + \phi)$ for some phase angle $\phi$ . To find $\phi$ , we use: …

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Q.The maximum distance of a point on the graph of the function y=3 sin⁡x+cos⁡xy = \sqrt{3}\,\sin x + \cos xy=3​sinx+cosx from xxx-axis is ______.

Why combine sine and cosine?

Solution

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Q.The maximum distance of a point on the graph of the function $y = \sqrt{3}\,\sin x + \cos x$ from $x$ -axis is ______.

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