Finding the Value of Angle θ in a Linear Transformation

The problem provides us with expressions for two variables, \(p\) and \(q\), in terms of other variables \(X\), \(Y\), and an angle \(\theta\). These expressions represent a rotation of coordinates. We are also given a quadratic relationship involving \(p\) and \(q\) which is equivalent to a quadratic relationship in \(X\) and \(Y\). Our goal is to determine the value of \(\theta\).

Understanding the Given Equations

We are given:

\( p = X \cos \theta - Y \sin \theta \)

\( q = X \sin \theta + Y \cos \theta \)

\( p^2 + 4pq + q^2 = AX^2 + BY^2 \)

The first two equations relate \((p, q)\) to \((X, Y)\) via a rotation by angle \(\theta\). The third equation gives a specific form that the combination \(p^2 + 4pq + q^2\) takes in terms of \(X\) and \(Y\).

Substituting p and q into the Quadratic Relation

We need to substitute the expressions for \(p\) and \(q\) into the equation \(p^2 + 4pq + q^2 = AX^2 + BY^2\). Let's calculate each term separately:

Calculate \(p^2\):

\[ p^2 = (X \cos \theta - Y \sin \theta)^2 = X^2 \cos^2 \theta - 2XY \cos \theta \sin \theta + Y^2 \sin^2 \theta \]

Calculate \(q^2\):

\[ q^2 = (X \sin \theta + Y \cos \theta)^2 = X^2 \sin^2 \theta + 2XY \sin \theta \cos \theta + Y^2 \cos^2 \theta \]

Calculate \(pq\):

\[ pq = (X \cos \theta - Y \sin \theta)(X \sin \theta + Y \cos \theta) \]

Expanding the product:

\[ pq = X^2 \cos \theta \sin \theta + XY \cos^2 \theta - Y^2 \sin \theta \cos \theta - XY \sin^2 \theta \]

Group terms with \(XY\):

\[ pq = XY (\cos^2 \theta - \sin^2 \theta) + X^2 \cos \theta \sin \theta - Y^2 \sin \theta \cos \theta \]

Using the double angle identities \( \cos(2\theta) = \cos^2 \theta - \sin^2 \theta \) and \( \sin(2\theta) = 2 \sin \theta \cos \theta \), we can rewrite \(pq\):

\[ pq = XY \cos(2\theta) + \frac{1}{2} X^2 \sin(2\theta) - \frac{1}{2} Y^2 \sin(2\theta) \]

Now substitute these into \(p^2 + 4pq + q^2\):

First, add \(p^2\) and \(q^2\):

\[ p^2 + q^2 = (X^2 \cos^2 \theta - 2XY \cos \theta \sin \theta + Y^2 \sin^2 \theta) + (X^2 \sin^2 \theta + 2XY \sin \theta \cos \theta + Y^2 \cos^2 \theta) \]

Notice that the \(XY\) terms cancel out:

\[ p^2 + q^2 = X^2 (\cos^2 \theta + \sin^2 \theta) + Y^2 (\sin^2 \theta + \cos^2 \theta) \]

Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):

\[ p^2 + q^2 = X^2(1) + Y^2(1) = X^2 + Y^2 \]

Now add \(4pq\):

\[ 4pq = 4 \left[ XY \cos(2\theta) + \frac{1}{2} X^2 \sin(2\theta) - \frac{1}{2} Y^2 \sin(2\theta) \right] \]

\[ 4pq = 4XY \cos(2\theta) + 2X^2 \sin(2\theta) - 2Y^2 \sin(2\theta) \]

So, \(p^2 + 4pq + q^2\) becomes:

\[ (X^2 + Y^2) + (4XY \cos(2\theta) + 2X^2 \sin(2\theta) - 2Y^2 \sin(2\theta)) \]

Group terms by \(X^2\), \(Y^2\), and \(XY\):

\[ X^2 (1 + 2 \sin(2\theta)) + Y^2 (1 - 2 \sin(2\theta)) + XY (4 \cos(2\theta)) \]

Equating Coefficients

We are given that \( p^2 + 4pq + q^2 = AX^2 + BY^2 \). So we have:

\[ X^2 (1 + 2 \sin(2\theta)) + Y^2 (1 - 2 \sin(2\theta)) + XY (4 \cos(2\theta)) = AX^2 + BY^2 \]

For this equation to hold true for all possible values of \(X\) and \(Y\), the coefficients of the corresponding terms on both sides must be equal.

Coefficient of \(X^2\): \( A = 1 + 2 \sin(2\theta) \)

Coefficient of \(Y^2\): \( B = 1 - 2 \sin(2\theta) \)

Coefficient of \(XY\): \( 4 \cos(2\theta) = 0 \)

Solving for θ

From the coefficient of the \(XY\) term, we have the equation:

\[ 4 \cos(2\theta) = 0 \]

This simplifies to:

\[ \cos(2\theta) = 0 \]

We are given the constraint \( 0 \le \theta \le \pi/2 \). Let's find the corresponding range for \(2\theta\):

Multiplying the inequality by 2:

\[ 0 \times 2 \le \theta \times 2 \le \frac{\pi}{2} \times 2 \]

\[ 0 \le 2\theta \le \pi \]

We need to find the value(s) of \(2\theta\) in the interval \( [0, \pi] \) for which the cosine is 0. The cosine function is zero at \( \pi/2 \).

So, we have:

\[ 2\theta = \frac{\pi}{2} \]

Solving for \(\theta\):

\[ \theta = \frac{(\pi/2)}{2} \]

\[ \theta = \frac{\pi}{4} \]

This value \(\theta = \pi/4\) lies within the given range \( 0 \le \theta \le \pi/2 \).

Conclusion

By substituting the expressions for \(p\) and \(q\) into the given equation and comparing the coefficients of \(X^2\), \(Y^2\), and \(XY\), we found that the coefficient of the \(XY\) term must be zero, which led to the trigonometric equation \(\cos(2\theta) = 0\). Solving this equation within the specified range for \(\theta\) gives the value \(\theta = \pi/4\).

The value of θ is \(\pi/4\).

Revision Table: Key Concepts

Additional Information: Quadratic Forms

The equation \( p^2 + 4pq + q^2 = AX^2 + BY^2 \) relates two quadratic forms. A quadratic form is a polynomial where every term has a total degree of two (like \(X^2\), \(Y^2\), \(XY\), \(p^2\), \(pq\), \(q^2\)).

The expression \( p^2 + 4pq + q^2 \) is a quadratic form in variables \(p\) and \(q\). The expression \( AX^2 + BY^2 \) is a quadratic form in variables \(X\) and \(Y\).

The fact that the quadratic form in \((p, q)\) transforms into a form \( AX^2 + BY^2 \) which contains only \(X^2\) and \(Y^2\) terms (no \(XY\) term) in the original coordinates \((X, Y)\) indicates that the chosen transformation angle \(\theta\) diagonalizes the quadratic form \( p^2 + 4pq + q^2 \) when expressed in \((X, Y)\) coordinates. The elimination of the \(XY\) term is key here, which is why setting its coefficient to zero was essential.

In general, a quadratic form \( ax^2 + bxy + cy^2 \) can be transformed into a form \( a'u^2 + c'v^2 \) (where the \(uv\) term is zero) by rotating the coordinate axes by a specific angle. This angle \(\theta\) is related to the coefficients \(a, b, c\).

In our problem, when \(p\) and \(q\) are substituted, the quadratic form \( p^2 + 4pq + q^2 \) in \(p,q\) variables becomes a quadratic form in \(X, Y\) variables. The fact that the result is given as \( AX^2 + BY^2 \) means the mixed term \(XY\) in the \(X, Y\) coordinates must vanish. The coefficient we calculated for the \(XY\) term, \(4 \cos(2\theta)\), being zero is exactly what is needed to eliminate this term and match the form \( AX^2 + BY^2 \).