Exercise 3.2 · Q13: Show $F(x)F(y)=F(x+y)$

Basics you need

Solution

1. Write down $F(y)$ and $F(x+y)$⁠; check the product is defined

   Replacing $x$ by $y$⁠, and then by $x + y$⁠:

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

   $F(x)$ is $3\times3$ and $F(y)$ is $3\times3$⁠. The number of columns of $F(x)$ (3) equals the number of rows of $F(y)$ (3), so $F(x)F(y)$ is defined, and it is $3\times3$⁠, the same order as $F(x+y)$⁠.

   Writing $F(x+y)$ out first shows the target: we must get $\cos(x+y)$⁠, $-\sin(x+y)$⁠, $\sin(x+y)$⁠, $\cos(x+y)$ in the top-left $2\times2$ block and the same 0s and 1 elsewhere.
   A product $AB$ needs (columns of $A$⁠) = (rows of $B$⁠). Lesson: matrix multiplication

2. Row 1 of the product

   Row 1 of $F(x)$ is $(\cos x,\ -\sin x,\ 0)$⁠. The columns of $F(y)$ are $(\cos y, \sin y, 0)$⁠, $(-\sin y, \cos y, 0)$ and $(0, 0, 1)$⁠, read top to bottom. Call the product $C = [c_{ik}]$⁠.

   $$\begin{aligned} c_{11} &= \cos x\cos y + (-\sin x)\sin y + 0\cdot 0\\ &= \cos x\cos y - \sin x\sin y \end{aligned}$$ $$\begin{aligned} c_{12} &= \cos x\,(-\sin y) + (-\sin x)\cos y\\ &\quad + 0\cdot 0\\ &= -(\sin x\cos y + \cos x\sin y) \end{aligned}$$ $$c_{13} = \cos x\cdot 0 + (-\sin x)\cdot 0 + 0\cdot 1 = 0$$
   Each entry pairs the three numbers of row 1 with the three numbers of one column, multiplies each pair, and adds. In $c_{12}$ both products are negative, so we take out a common factor $-1$⁠; the bracket left over is in the order of the sine formula.
   $c_{ik} = a_{i1}b_{1k} + a_{i2}b_{2k} + a_{i3}b_{3k}$⁠. Basics: pair up, multiply, add

3. Row 2 of the product

   Row 2 of $F(x)$ is $(\sin x,\ \cos x,\ 0)$⁠. With the same three columns:

   $$\begin{aligned} c_{21} &= \sin x\cos y + \cos x\sin y + 0\cdot 0\\ &= \sin x\cos y + \cos x\sin y \end{aligned}$$ $$\begin{aligned} c_{22} &= \sin x\,(-\sin y) + \cos x\cos y + 0\cdot 0\\ &= \cos x\cos y - \sin x\sin y \end{aligned}$$ $$c_{23} = \sin x\cdot 0 + \cos x\cdot 0 + 0\cdot 1 = 0$$
   In $c_{22}$ we reorder the two terms (addition of numbers is commutative) so that it matches the cosine formula exactly.

4. Row 3 of the product

   Row 3 of $F(x)$ is $(0,\ 0,\ 1)$⁠:

   $$\begin{aligned} c_{31} &= 0\cdot\cos y + 0\cdot\sin y + 1\cdot 0 = 0\\ c_{32} &= 0\cdot(-\sin y) + 0\cdot\cos y + 1\cdot 0 = 0\\ c_{33} &= 0\cdot 0 + 0\cdot 0 + 1\cdot 1 = 1 \end{aligned}$$
   The two zeros in row 3 kill the first two products every time, so only the third entry of each column survives: 0, 0, 1.

5. Apply the compound angle formulas

   For all real $x$ and $y$⁠,

   $$\begin{aligned} \cos x\cos y - \sin x\sin y &= \cos(x+y),\\ \sin x\cos y + \cos x\sin y &= \sin(x+y). \end{aligned}$$

   So, from Steps 2 and 3,

   $$\begin{aligned} c_{11} &= \cos(x+y), \\ c_{12} &= -\sin(x+y),\\ c_{21} &= \sin(x+y), \\ c_{22} &= \cos(x+y). \end{aligned}$$
   These identities hold for every pair of real numbers, with no restriction on $x$ or $y$⁠, so the argument works for all $x$ and $y$⁠, not just for particular angles.
   Compound angle formulas (Class XI, Chapter 3). Basics: compound angles

6. Assemble and compare

   Putting the nine entries in place,

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

   This has the same order as $F(x+y)$ from Step 1, and every entry agrees with the entry in the same position. Hence $F(x)F(y) = F(x+y)$.

   Two matrices are equal when they have the same order and equal corresponding entries (Definition 2). We have checked all nine positions.

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