IBDP Maths AI: Topic : AHL 1.14: Definition of a matrix: IB style Questions HL Paper 1

Detailed Solution

Part (a): Given Rotation Matrix \( R(2\alpha) \)
The matrix is:

\[ R(2\alpha) = \begin{pmatrix} \cos(2\alpha) & -\sin(2\alpha) \\ \sin(2\alpha) & \cos(2\alpha) \end{pmatrix} \]

This is the standard form of a 2D rotation matrix for an angle \( 2\alpha \), which rotates a vector counterclockwise by \( 2\alpha \) radians. No computation is needed here; it’s provided as a reference.

Part (b): Compute \( R(\alpha) \times R(\alpha) \)
We’re given:

\[ R(\alpha) = \begin{pmatrix} \cos(\alpha) & -\sin(\alpha) \\ \sin(\alpha) & \cos(\alpha) \end{pmatrix} \]

We need to verify the matrix multiplication:

\[ R(\alpha) \times R(\alpha) = \begin{pmatrix} \cos(\alpha) & -\sin(\alpha) \\ \sin(\alpha) & \cos(\alpha) \end{pmatrix} \times \begin{pmatrix} \cos(\alpha) & -\sin(\alpha) \\ \sin(\alpha) & \cos(\alpha) \end{pmatrix} \]

Perform the multiplication:

Element (1,1):
\[ \cos(\alpha) \cdot \cos(\alpha) + (-\sin(\alpha)) \cdot \sin(\alpha) = \cos^2(\alpha) – \sin^2(\alpha) \]

Element (1,2):
\[ \cos(\alpha) \cdot (-\sin(\alpha)) + (-\sin(\alpha)) \cdot \cos(\alpha) = -\cos(\alpha)\sin(\alpha) – \sin(\alpha)\cos(\alpha) = -2\sin(\alpha)\cos(\alpha) \]

Element (2,1):
\[ \sin(\alpha) \cdot \cos(\alpha) + \cos(\alpha) \cdot \sin(\alpha) = \sin(\alpha)\cos(\alpha) + \cos(\alpha)\sin(\alpha) = 2\sin(\alpha)\cos(\alpha) \]

Element (2,2):
\[ \sin(\alpha) \cdot (-\sin(\alpha)) + \cos(\alpha) \cdot \cos(\alpha) = -\sin^2(\alpha) + \cos^2(\alpha) = \cos^2(\alpha) – \sin^2(\alpha) \]

Resulting matrix:

\[ R(\alpha) \times R(\alpha) = \begin{pmatrix} \cos^2(\alpha) – \sin^2(\alpha) & -2\sin(\alpha)\cos(\alpha) \\ 2\sin(\alpha)\cos(\alpha) & \cos^2(\alpha) – \sin^2(\alpha) \end{pmatrix} \]

This matches exactly what you provided in part (b), confirming the multiplication is correct. Geometrically, multiplying two rotation matrices \( R(\alpha) \) represents a rotation by \( \alpha + \alpha = 2\alpha \), so we expect \( R(\alpha) \times R(\alpha) = R(2\alpha) \), which we’ll verify via part (c).

Part (c): Derive Trigonometric Identities

(i) First Row, Second Column Comparison
Compare:

From \( R(2\alpha) \): Element (1,2) = \( -\sin(2\alpha) \)
From \( R(\alpha) \times R(\alpha) \): Element (1,2) = \( -2\sin(\alpha)\cos(\alpha) \)

Equate them:

\[ -\sin(2\alpha) = -2\sin(\alpha)\cos(\alpha) \]

Multiply both sides by \(-1\):

\[ \sin(2\alpha) = 2\sin(\alpha)\cos(\alpha) \]

This is the standard double-angle identity for sine, which is correct and confirms that the matrix multiplication aligns with the rotation property \( R(\alpha) \times R(\alpha) = R(2\alpha) \).

(ii) First Row, First Column Comparison
Compare:

– From \( R(2\alpha) \): Element (1,1) = \( \cos(2\alpha) \)
– From \( R(\alpha) \times R(\alpha) \): Element (1,1) = \( \cos^2(\alpha) – \sin^2(\alpha) \)

Equate them:

\[ \cos(2\alpha) = \cos^2(\alpha) – \sin^2(\alpha) \]

This is a known double-angle identity for cosine. To derive the alternative form using \( \sin^2(\alpha) + \cos^2(\alpha) = 1 \):

\[ \cos^2(\alpha) – \sin^2(\alpha) = \cos^2(\alpha) – (1 – \cos^2(\alpha)) = \cos^2(\alpha) – 1 + \cos^2(\alpha) = 2\cos^2(\alpha) – 1 \]

Alternatively:

\[ \cos^2(\alpha) – \sin^2(\alpha) = (1 – \sin^2(\alpha)) – \sin^2(\alpha) = 1 – 2\sin^2(\alpha) \]

So:

\[ \cos(2\alpha) = 1 – 2\sin^2(\alpha) \]

This matches your derivation and is another standard form of the cosine double-angle identity. Both forms (\( \cos^2(\alpha) – \sin^2(\alpha) \) and \( 1 – 2\sin^2(\alpha) \)) are equivalent and correct.

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Solution: –

(a) $R(2\alpha)$ = \begin{pmatrix} \cos(2\alpha) & -\sin(2\alpha) \\ \sin(2\alpha) & \cos(2\alpha) \end{pmatrix}

(b) $R(\alpha)$ $R(\alpha)$  = \begin{pmatrix} \cos(\alpha) & -\sin(\alpha) \\ \sin(\alpha) & \cos(\alpha) \end{pmatrix} \begin{pmatrix} \cos(\alpha) & -\sin(\alpha) \\ \sin(\alpha) & \cos(\alpha) \end{pmatrix}
which is =  \begin{pmatrix} \cos^2(\alpha)-\sin^2(\alpha) & -2\cos(\alpha)\sin(\alpha) \\ 2\cos(\alpha)\sin(\alpha) & \cos^2(\alpha)-\sin^2(\alpha) \end{pmatrix}

(c) (i) (Because matrix multiplication represents the composition of transformations, two rotations of $\alpha$ are equivalent to a rotation of $2\alpha$, so the two matrices are equal (so each of their corresponding elements are equal))
$\sin(2\alpha) = 2\sin(\alpha)\cos(\alpha)$

(ii) $\cos(2\alpha) = \cos^2(\alpha)-\sin^2(\alpha)$
replacing $\cos^2(\alpha)$ with $1-\sin^2(\alpha)$
$= 1-\sin^2(\alpha)-\sin^2(\alpha)$
$= 1-2\sin^2(\alpha)$