If $F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$, show that $F(x) F(y) = F(x+y)$.

If F(x) = cos x -sin x 0 ; sin x cos x 0 \0 0 1 , show that F(x) F(y) = F(x+y).

Exercise 3.2 · Q13

F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}

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Appeared in past exams:CBSE 2023· Set 65/2/1· 5mreworded

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Concept understanding — Rotation Matrix Composition

Rotation Matrix Composition

Hold a book, rotate it so the cover faces you, then tilt it to look down at the spine. The final orientation depends on which turn you do first. That is the whole idea: composing rotations means applying one after another, and in 3D the order matters.

Each rotation is a matrix, and doing two in sequence corresponds to multiplying their matrices. The subtlety is that matrix multiplication is generally not commutative: $AB \neq BA$ .

The Rule

R_1

Read right to left: a rotation acts on a column vector as $R\mathbf{v}$ , so applying $R_1$ then $R_2$ gives $R_2(R_1\mathbf{v}) = (R_2 R_1)\mathbf{v}$ . The rotation done first sits on the right.

A 2D Example

In the plane, a counter-clockwise rotation by angle $\theta$ is

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

Here $R(60^\circ)\,R(30^\circ) = R(90^\circ)$ : the angles simply add. In 2D rotations do commute, because they all share the same axis (the axis pointing out of the plane).

Why 3D Is Different

In 3D, rotations are about different axes, and swapping the order changes the result. Use the standard right-handed matrices

R_x(90^\circ) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}, \qquad R_y(90^\circ) = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ -1 & 0 & 0 \end{pmatrix},

and track the vector $\mathbf{v} = (0,0,1)$ .

Order 1 — $R_x$ first, then $R_y$ (product $R_y R_x$ ):

$R_x(90^\circ)\mathbf{v} = (0,-1,0)$ — the point moves onto the $y$ -axis.
$R_y$ leaves the $y$ -axis fixed, so the result stays $(0,-1,0)$ .

Order 2 — $R_y$ first, then $R_x$ (product $R_x R_y$ ):

$R_y(90^\circ)\mathbf{v} = (1,0,0)$ — the point moves onto the $x$ -axis.
$R_x$ leaves the $x$ -axis fixed, so the result stays $(1,0,0)$ .

Watch out

Same two rotations, opposite order, different final point: $(0,-1,0)$ versus $(1,0,0)$ . So $R_x R_y \neq R_y R_x$ in general. (Notice each rotation leaves its own axis unmoved — a rotation about the $y$ -axis can never shift a vector that already lies on the $y$ -axis.) …

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The matrix $F(x)$ represents a rotation by angle $x$ about the $z$ -axis in 3D space. Multiplying two such rotations corresponds to adding the angles, so $F(x)F(y) = F(x+y)$ — a direct consequence of the angle‑addition formulas for sine and cosine.

Why this works

The matrix $F(x)$ is the standard rotation matrix for a counter‑clockwise rotation by angle $x$ around the $z$ ‑axis. In 3D, rotating by $x$ and then by $y$ is the same as rotating by $x+y$ in one step. The algebra must reflect this geometric fact — and it does, because the product of two rotation matrices is another rotation matrix whose angle is the sum.

The key tools are the trigonometric identities:

\cos(x+y) = \cos x \cos y - \sin x \sin y

\sin(x+y) = \sin x \cos y + \cos x \sin y

Step‑by‑step verification

1. Write down $F(x)$ and $F(y)$ explicitly.

F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}, \quad F(y) = \begin{bmatrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix}

2. Multiply $F(x)F(y)$ .

Only the top‑left $2\times2$ block is non‑trivial; the third row and column are just $[0\;0\;1]$ . So we focus on the $2\times2$ product:

\begin{bmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{bmatrix} \begin{bmatrix} \cos y & -\sin y \\ \sin y & \cos y \end{bmatrix}

Compute entry by entry:

Top‑left: $\cos x \cos y + (-\sin x)(\sin y) = \cos x \cos y - \sin x \sin y = \cos(x+y)$
Top‑right: $\cos x (-\sin y) + (-\sin x)(\cos y) = -\cos x \sin y - \sin x \cos y = -\sin(x+y)$
Bottom‑left: $\sin x \cos y + \cos x \sin y = \sin(x+y)$
Bottom‑right: $\sin x (-\sin y) + \cos x \cos y = -\sin x \sin y + \cos x \cos y = \cos(x+y)$ …

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

AHSEC Higher Secondary (HS) Final Examination 2023Set ANNUAL4 marks
$Q. Answer$
(i) and (ii):
(i) If $F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$ , show that $F(x)F(y) = F(x+y)$ .
(ii) Prove that $\displaystyle\int_0^{2a} f(x)\,dx = \int_0^a f(x)\,dx + \int_0^a f(2a-x)\,dx$ . OR Answer
(a) and (b):
(a) If $\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}$ , then find $x$ .
(b) If $x = a(\cos t + t\sin t)$ , $y = a(\sin t - t\cos t)$ , find $\dfrac{dy}{dx}$ .
›Reveal solutionSolution

(i) Matrix multiplication + angle-sum identities; (ii) split the integral and substitute $x=2a-t$ in the second piece; (OR-a) equate determinants; (OR-b) differentiate the parametric equations and divide.

(i) $F(x) = \begin{bmatrix}\cos x & -\sin x & 0\\ \sin x & \cos x & 0\\ 0&0&1\end{bmatrix}$ . Compute $F(x)F(y)$ by matrix multiplication:

Row 1: $\big[\cos x\cos y - \sin x\sin y,\ -\cos x\sin y - \sin x\cos y,\ 0\big] = [\cos(x+y),\ -\sin(x+y),\ 0]$

Row 2: $\big[\sin x\cos y + \cos x\sin y,\ -\sin x\sin y + \cos x\cos y,\ 0\big] = [\sin(x+y),\ \cos(x+y),\ 0]$

Row 3: $[0,0,1]$

using $\cos x\cos y - \sin x\sin y = \cos(x+y)$ and $\sin x\cos y+\cos x\sin y = \sin(x+y)$ . This is exactly $F(x+y)$ . Hence $F(x)F(y) = F(x+y)$ .

(ii) $\int_0^{2a}f(x)dx = \int_0^a f(x)dx + \int_a^{2a}f(x)dx$ (splitting the interval).

In $\int_a^{2a}f(x)dx$ , substitute $x = 2a-t$ , so $dx=-dt$ ; when $x=a, t=a$ ; when $x=2a, t=0$ :

$\int_a^{2a}f(x)dx = \int_a^0 f(2a-t)(-dt) = \int_0^a f(2a-t)\,dt = \int_0^a f(2a-x)\,dx$ (renaming the dummy variable).

Therefore $\displaystyle\int_0^{2a}f(x)dx = \int_0^a f(x)dx + \int_0^a f(2a-x)dx$ , as required.

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Q.If $F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$ , show that $F(x) F(y) = F(x+y)$ .

Concept understanding — Rotation Matrix Composition

Rotation Matrix Composition

The Rule

A 2D Example

Why 3D Is Different

Why this works

Step‑by‑step verification

Method: Proving a matrix identity by direct multiplication and known formulas

Steps

Common Mistakes

Q.If F(x)=[cos⁡x−sin⁡x0sin⁡xcos⁡x0001]F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}F(x)=​cosxsinx0​−sinxcosx0​001​​, show that F(x)F(y)=F(x+y)F(x) F(y) = F(x+y)F(x)F(y)=F(x+y).

Why this works

Step‑by‑step verification

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Q.If $F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$ , show that $F(x) F(y) = F(x+y)$ .

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