The maximum value of $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 + \sin\theta & 1 \\ 1 + \cos\theta & 1 & 1 \end{vmatrix}$ is ($\theta$ is real number) (A) $\dfrac{1}{2}$ (B) $\dfrac{\sqrt{3}}{2}$ (C) $\sqrt{2}$ (D) $\dfrac{2\sqrt{3}}{4}$

The maximum value of Δ = 1 1 1 \1 1 + sinθ 1 \1 + cosθ 1 1 is (θ is real number) (A) (1)/(2) (B) dfrac√(3)2 (C

NCERT Exemplar · Q31

\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 + \sin\theta & 1 \\ 1 + \cos\theta & 1 & 1 \end{vmatrix}

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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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$\Delta = -\sin\theta\cos\theta = -\tfrac12\sin 2\theta$ , whose maximum value is $\tfrac12$ . Correct option: (A).

Apply the row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$ (these do not change the value of the determinant):

\Delta = \begin{vmatrix} 1 & 1 & 1 \\0 & \sin\theta & 0 \\\cos\theta & 0 & 0 \end{vmatrix}.

Expand along the third row, which has two zeros:

$\Delta = \cos\theta\,(+1)\begin{vmatrix} 1 & 1 \\\sin\theta & 0 \end{vmatrix} = \cos\theta\,(0 - \sin\theta) = -\sin\theta\cos\theta.$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2026Set 2026-A1 markMCQ
$f(x)=\sin x+\sqrt{3} \cos x$
›Reveal solutionSolution
The function $f(x)=\sin x+\sqrt{3}\cos x$ can be rewritten as $2\sin(x+\pi/3)$ . On $[0,\pi]$ , the maximum is $2$ and the minimum is $-\sqrt{3}$ , so the correct option is (D).

Concept & Intuition

When a function is a linear combination of $\sin x$ and $\cos x$ , we can combine them into a single sine (or cosine) with a phase shift. This is because

$A\sin x + B\cos x = R\sin(x+\phi)$ where $R=\sqrt{A^2+B^2}$ and $\phi$ satisfies $\cos\phi = A/R$ , $\sin\phi = B/R$ .

Here $A=1$ , $B=\sqrt{3}$ , so $R=2$ and $\phi=\pi/3$ . Then the problem reduces to finding the extreme values of $2\sin(x+\pi/3)$ on $x\in[0,\pi]$ , which is straightforward.

Step-by-step solution

Rewrite the function

$f(x)=\sin x+\sqrt{3}\cos x$ .

Compute $R=\sqrt{1^2+(\sqrt{3})^2}=\sqrt{1+3}=2$ .

Find $\phi$ such that $\cos\phi = \frac{1}{2}$ , $\sin\phi = \frac{\sqrt{3}}{2}$ . This gives $\phi = \frac{\pi}{3}$ .

Hence $f(x)=2\sin\left(x+\frac{\pi}{3}\right)$ .

Determine the range of the argument

Since $x\in[0,\pi]$ , then $x+\frac{\pi}{3}\in\left[\frac{\pi}{3},\,\frac{4\pi}{3}\right]$ .

Find the maximum

The sine function attains its maximum value $1$ at $\frac{\pi}{2}$ .

Check if $\frac{\pi}{2}$ lies in $\left[\frac{\pi}{3},\frac{4\pi}{3}\right]$ : yes, it does.

So the maximum of $f$ is $2\cdot 1 = 2$ .

Find the minimum

On $\left[\frac{\pi}{3},\frac{4\pi}{3}\right]$ , sine decreases from $\sin(\pi/3)=\sqrt{3}/2$ to $\sin(\pi)=0$ , then continues to $\sin(4\pi/3)=-\sqrt{3}/2$ . …
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KCET 2020Set A-11 markMCQ
$\sin^2 51° + \sin^2 39°$
›Reveal solutionSolution

Spot that the two angles are complementary, convert one sine into a cosine, and finish with $\sin^2\theta+\cos^2\theta=1$ .

Step 1 — Notice the complementary pair.
$51^\circ+39^\circ=90^\circ .$
Whenever the angles in such an expression add to $90^\circ$ , the co-function identity is the intended route:
$\sin(90^\circ-\theta)=\cos\theta .$
Step 2 — Rewrite the second term.

Taking $\theta=51^\circ$ :
$\sin 39^\circ=\sin(90^\circ-51^\circ)=\cos 51^\circ\ \Rightarrow\ \sin^2 39^\circ=\cos^2 51^\circ .$
Step 3 — Apply the Pythagorean identity.

$\sin^2 51^\circ+\sin^2 39^\circ=\sin^2 51^\circ+\cos^2 51^\circ=1 .$ …
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KCET 2026Set UNKNOWN1 markMCQ
$\sin(x + \pi/6) + \cos(x + \pi/6)$
›Reveal solutionSolution

Combine the sine and cosine into a single sine using $\sin\theta + \cos\theta = \sqrt{2}\sin(\theta + \pi/4)$ , then find where that sine is maximum.

Step 1 — Combine into a single sinusoid

Let $\theta = x + \pi/6$ . Using $\sin\theta + \cos\theta = \sqrt{2}\sin\left(\theta + \dfrac{\pi}{4}\right)$ :
$\sin(x+\pi/6) + \cos(x+\pi/6) = \sqrt{2}\,\sin\!\left(x + \frac{\pi}{6} + \frac{\pi}{4}\right)$
Step 2 — Maximize

$\sqrt{2}\sin(\cdot)$ attains its maximum value $\sqrt{2}$ when the angle inside equals $\pi/2$ :
$x + \frac{\pi}{6} + \frac{\pi}{4} = \frac{\pi}{2}$
Step 3 — Solve for $x$ …
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Q.The maximum value of $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 + \sin\theta & 1 \\ 1 + \cos\theta & 1 & 1 \end{vmatrix}$ is ( $\theta$ is real number)
(A) $\dfrac{1}{2}$
(B) $\dfrac{\sqrt{3}}{2}$
(C) $\sqrt{2}$
(D) $\dfrac{2\sqrt{3}}{4}$

Concept understanding — Maximum Value Sine Cosine

Maximum Value of Sine and Cosine – The Core Idea

The Precise Statement

Why This Matters in Exams

Common Mistake to Avoid …

Method: Simplifying a Determinant to Bound Its Maximum or Minimum Value

Steps

Common Mistakes

Step 1 — Combine into a single sinusoid

Step 2 — Maximize

Step 3 — Solve for $x$ …

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