Trigonometric Matrices: JEE's Hidden Favorites

Introduction

Matrices with $\sin\theta$, $\cos\theta$, and related trigonometric functions appear repeatedly in JEE. These matrices often have elegant properties that make calculations surprisingly simple — if you recognize them!

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Part I: The Rotation Matrix (Most Important!)

1.1 Definition

The 2D Rotation Matrix by angle $\theta$ (counterclockwise):

$R(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$

1.2 Fundamental Properties

1.3 The Golden Property: Powers

$\boxed{R(\theta)^n = R(n\theta)}$

Proof: $R(\theta)^2 = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}^2 = \begin{pmatrix} \cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{pmatrix} = R(2\theta)$

By induction: $R(\theta)^n = R(n\theta)$

1.4 Composition Property

$R(\alpha) \cdot R(\beta) = R(\alpha + \beta)$

This is essentially the angle addition formula in matrix form!

1.5 Special Values

1.6 Eigenvalues

$\lambda = \cos\theta \pm i\sin\theta = e^{\pm i\theta}$

Both eigenvalues lie on the unit circle in the complex plane.

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Part II: The Reflection Matrix

2.1 Definition

Reflection across a line making angle $\theta$ with the positive x-axis:

$M(\theta) = \begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}$

2.2 Fundamental Properties

2.3 Powers

Since $M^2 = I$: $M^n = \begin{cases} I & \text{if } n \text{ even} \\ M & \text{if } n \text{ odd} \end{cases}$

2.4 Special Cases

2.5 Relation to Rotation

$M(\theta) = R(2\theta) \cdot M(0)$

Reflection = Rotation followed by reflection across x-axis.

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Part III: The 90° Rotation Matrix (Special Case)

3.1 Definition

$J = R\left(\frac{\pi}{2}\right) = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$

3.2 Remarkable Properties

3.3 Cyclic Pattern

$J^n = \begin{cases} I & n \equiv 0 \pmod{4} \\ J & n \equiv 1 \pmod{4} \\ -I & n \equiv 2 \pmod{4} \\ -J & n \equiv 3 \pmod{4} \end{cases}$

3.4 The "Matrix i"

$J$ behaves like the imaginary unit $i$ in matrix form:

• $J^2 = -I$ (like $i^2 = -1$)
• Any matrix $aI + bJ$ behaves like complex number $a + bi$

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Part IV: Projection Matrices

4.1 Projection onto a Line

Projection onto line making angle $\theta$ with x-axis:

$P(\theta) = \begin{pmatrix} \cos^2\theta & \sin\theta\cos\theta \\ \sin\theta\cos\theta & \sin^2\theta \end{pmatrix}$

4.2 Properties

4.3 Complement Projection

Projection onto the perpendicular line: $P_\perp(\theta) = I - P(\theta) = P\left(\theta + \frac{\pi}{2}\right)$

4.4 Relation to Reflection

$M(\theta) = 2P(\theta) - I$

Or equivalently: $P(\theta) = \frac{I + M(\theta)}{2}$

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Part V: The Symmetric Trig Matrix

5.1 Common Form

$A = \begin{pmatrix} 1 + \sin^2\theta & \sin\theta\cos\theta \\ \sin\theta\cos\theta & 1 + \cos^2\theta \end{pmatrix}$

5.2 Properties

$|A| = (1 + \sin^2\theta)(1 + \cos^2\theta) - \sin^2\theta\cos^2\theta$ $= 1 + \sin^2\theta + \cos^2\theta + \sin^2\theta\cos^2\theta - \sin^2\theta\cos^2\theta$ $= 1 + 1 = 2$

Always a constant! Independent of $\theta$.

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Part VI: 3×3 Rotation Matrices

6.1 Rotation about x-axis

$R_x(\theta) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos\theta & -\sin\theta \\ 0 & \sin\theta & \cos\theta \end{pmatrix}$

6.2 Rotation about y-axis

$R_y(\theta) = \begin{pmatrix} \cos\theta & 0 & \sin\theta \\ 0 & 1 & 0 \\ -\sin\theta & 0 & \cos\theta \end{pmatrix}$

6.3 Rotation about z-axis

$R_z(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix}$

6.4 Common Properties

All 3D rotation matrices satisfy:

• $|R| = 1$
• $R^T = R^{-1}$
• $R^n(\theta) = R(n\theta)$
• Eigenvalues: $1, e^{i\theta}, e^{-i\theta}$

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Part VII: Determinant Patterns with Trig Elements

7.1 The Classic Determinant

$\begin{vmatrix} 1 & \cos\alpha & \cos\beta \\ \cos\alpha & 1 & \cos\gamma \\ \cos\beta & \cos\gamma & 1 \end{vmatrix}$

This equals zero when $\alpha, \beta, \gamma$ are angles of a triangle (since $\alpha + \beta + \gamma = \pi$).

7.2 Vandermonde-like Trig Determinant

$\begin{vmatrix} 1 & \sin\alpha & \cos\alpha \\ 1 & \sin\beta & \cos\beta \\ 1 & \sin\gamma & \cos\gamma \end{vmatrix}$

7.3 The Magic Determinant

$\begin{vmatrix} \cos(A-B) & \cos(B-C) & \cos(C-A) \\ \cos(A+B) & \cos(B+C) & \cos(C+A) \\ \sin(A+B) & \sin(B+C) & \sin(C+A) \end{vmatrix} = 0$

This is always zero regardless of $A, B, C$!

7.4 sin²/cos² Determinants

$\begin{vmatrix} \sin^2 A & \cot A & 1 \\ \sin^2 B & \cot B & 1 \\ \sin^2 C & \cot C & 1 \end{vmatrix} = 0 \text{ when } A + B + C = \pi$

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Part VIII: JEE Previous Year Questions

PYQ 1: JEE Main 2018

Problem: Let $A = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$. Find $A^{50}$ when $\theta = \frac{\pi}{12}$.

Solution:

Using $R(\theta)^n = R(n\theta)$:

$A^{50} = R\left(\frac{50\pi}{12}\right) = R\left(\frac{25\pi}{6}\right)$

Reduce the angle: $\frac{25\pi}{6} = 4\pi + \frac{\pi}{6} = \frac{\pi}{6} \pmod{2\pi}$

$A^{50} = R\left(\frac{\pi}{6}\right) = \begin{pmatrix} \cos\frac{\pi}{6} & -\sin\frac{\pi}{6} \\ \sin\frac{\pi}{6} & \cos\frac{\pi}{6} \end{pmatrix} = \boxed{\begin{pmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix}}$

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PYQ 2: JEE Main 2020

Problem: If $A = \begin{pmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{pmatrix}$ and $A + A^T = I$, find $\alpha$.

Solution:

$A + A^T = \begin{pmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{pmatrix} + \begin{pmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{pmatrix}$

$= \begin{pmatrix} 2\cos\alpha & 0 \\ 0 & 2\cos\alpha \end{pmatrix} = 2\cos\alpha \cdot I$

Given: $A + A^T = I$

$2\cos\alpha = 1 \implies \cos\alpha = \frac{1}{2}$

$\alpha = \boxed{\frac{\pi}{3}} \text{ (principal value)}$

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PYQ 3: JEE Main 2019

Problem: If $A = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}$, then find the value of $\theta$ for which $A$ is an identity matrix.

Solution:

For $A = I$: $\cos\theta = 1 \text{ and } \sin\theta = 0$

$\theta = 2n\pi, \quad n \in \mathbb{Z}$

Principal value: $\boxed{\theta = 0}$

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PYQ 4: JEE Advanced 2016

Problem: Let $P = \begin{pmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix}$, $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$, and $Q = PAP^T$. If $P^TQ^{2023}P = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, find $a + b + c + d$.

Solution:

First, recognize $P = R\left(-\frac{\pi}{6}\right)$ (rotation by $-30°$).

Note: $P^T = P^{-1} = R\left(\frac{\pi}{6}\right)$

$Q = PAP^T$

$P^TQ^nP = P^T(PAP^T)^nP$

Using the property $(PAP^T)^n = PA^nP^T$:

$P^TQ^nP = P^T \cdot PA^nP^T \cdot P = A^n$

So we need $A^{2023}$.

$A = I + N$ where $N = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, and $N^2 = O$.

$A^{2023} = (I + N)^{2023} = I + 2023N = \begin{pmatrix} 1 & 2023 \\ 0 & 1 \end{pmatrix}$

$a + b + c + d = 1 + 2023 + 0 + 1 = \boxed{2025}$

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PYQ 5: JEE Main 2021

Problem: If $A = \begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}$ and $A^5 = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, find $a^2 + b^2 + c^2 + d^2$.

Solution:

$A$ is a reflection matrix, so $A^2 = I$.

$A^5 = A^4 \cdot A = (A^2)^2 \cdot A = I^2 \cdot A = A$

So $\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}$

$a^2 + b^2 + c^2 + d^2 = \cos^2 2\theta + \sin^2 2\theta + \sin^2 2\theta + \cos^2 2\theta$ $= 1 + 1 = \boxed{2}$

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PYQ 6: JEE Main 2022

Problem: Evaluate: $\begin{vmatrix} \sin^2 x & \cos^2 x & 1 \\ \cos^2 x & \sin^2 x & 1 \\ -10 & 12 & 2 \end{vmatrix}$

Solution:

$C_1 + C_2$: First two columns add to give $\sin^2 x + \cos^2 x = 1$ in rows 1, 2.

$= \begin{vmatrix} \sin^2 x & 1 & 1 \\ \cos^2 x & 1 & 1 \\ -10 & 2 & 2 \end{vmatrix}$

Columns 2 and 3 are identical!

$\boxed{= 0}$

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PYQ 7: JEE Main 2023

Problem: If $f(\theta) = \begin{vmatrix} 1 & \sin\theta & 1 \\ -\sin\theta & 1 & \sin\theta \\ -1 & -\sin\theta & 1 \end{vmatrix}$, then $f(\theta)$ lies in the range:

Solution:

Expand along R1: $f(\theta) = 1(1 + \sin^2\theta) - \sin\theta(-\sin\theta + \sin\theta) + 1(\sin^2\theta + 1)$ $= (1 + \sin^2\theta) - 0 + (1 + \sin^2\theta)$ $= 2(1 + \sin^2\theta)$ $= 2 + 2\sin^2\theta$

Since $0 \leq \sin^2\theta \leq 1$: $2 \leq f(\theta) \leq 4$

Range: $\boxed{[2, 4]}$

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PYQ 8: JEE Advanced 2019

Problem: Let $\omega = e^{i\frac{2\pi}{3}}$ and $A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega^4 \end{pmatrix}$. Then $A^3$ equals:

Solution:

This is related to the DFT matrix. Note that $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.

First, find $A^2$ or observe the pattern.

For this specific matrix (scaled DFT matrix): $A \cdot \bar{A} = 3I$

After calculation: $A^2 = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 0 & 3 \\ 0 & 3 & 0 \end{pmatrix}$

$A^3 = A^2 \cdot A = 3 \cdot \begin{pmatrix} 1 & 1 & 1 \\ 1 & \omega^2 & \omega^4 \\ 1 & \omega & \omega^2 \end{pmatrix} = 3\bar{A}$

But more directly, for this DFT-type matrix: $A^3 = 3^{3/2} \cdot P = 3\sqrt{3} \cdot P$

After careful computation: $\boxed{A^3 = 3\bar{A}}$ or equivalently a specific matrix form.

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PYQ 9: JEE Main 2019

Problem: If $A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ (the 90° rotation matrix), find $(I + A)^{100}$.

Solution:

Note: $A = J$ where $J^2 = -I$, $J^4 = I$.

Let's compute $(I + A)^2$ first: $(I + A)^2 = I + 2A + A^2 = I + 2A - I = 2A$

$(I + A)^4 = (2A)^2 = 4A^2 = -4I$

$(I + A)^8 = (-4I)^2 = 16I$

$(I + A)^{16} = 256I$

$(I + A)^{32} = 256^2 I$

$(I + A)^{64} = 256^4 I$

$(I + A)^{96} = 256^6 I = 2^{48}I$

$(I + A)^{100} = (I + A)^{96} \cdot (I + A)^4 = 2^{48}I \cdot (-4I) = -2^{50}I$

$= \boxed{-2^{50}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}}$

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PYQ 10: JEE Main 2020

Problem: If the matrix $A = \begin{pmatrix} \cos^2\theta & \cos\theta\sin\theta \\ \cos\theta\sin\theta & \sin^2\theta \end{pmatrix}$ and $B = \begin{pmatrix} \cos^2\phi & \cos\phi\sin\phi \\ \cos\phi\sin\phi & \sin^2\phi \end{pmatrix}$, when is $AB = O$?

Solution:

Both $A$ and $B$ are projection matrices.

$A$ projects onto the line at angle $\theta$. $B$ projects onto the line at angle $\phi$.

$AB = O$ when the lines are perpendicular: $\theta - \phi = \pm\frac{\pi}{2}$

Or: $\boxed{\theta = \phi \pm \frac{\pi}{2}}$

Verification: When $\theta = 0$, $\phi = \frac{\pi}{2}$: $A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \quad B = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$ $AB = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = O \quad ✓$

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Part IX: Quick Reference Tables

Rotation Matrix $R(\theta)$

Reflection Matrix $M(\theta)$

90° Rotation $J$

Projection Matrix $P(\theta)$

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Part X: Recognition Patterns

How to Identify

Determinant Shortcuts

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Conclusion

Trigonometric matrices are elegant because:

1. Rotation matrices convert powers to simple angle multiplication
2. Reflection matrices are involutory — all high powers reduce to $I$ or $M$
3. Projection matrices are idempotent — all powers equal $P$
4. Determinants often simplify using $\sin^2 + \cos^2 = 1$

JEE Strategy:

• Recognize the matrix type FIRST
• Apply the appropriate power formula
• Use periodicity to reduce large exponents
• Look for trig identities hiding in determinants

Master these patterns, and trigonometric matrix problems become almost trivial!

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Last updated: January 2026 | Essential for JEE Main & Advanced