If `A=[{:(Costheta ,isintheta),("isintheta,costheta):}],(theta=pi/24)and A^(5)=[(a,b),(c,d)],` where `i=sqrt(-1)` then, which one of the following is not true ?

Options

A. `a^2-d^2=0`
B. `a^2-c^2=1`
C. `a^2-b^2=1/2`
D. `0lea^2+b^2le1`

Answer

To solve the problem, we start with the given matrix \( A \):

\[ A = \begin{pmatrix} \cos \theta & i \sin \theta \\ i \sin \theta & \cos \theta \end{pmatrix} \]

where \( \theta = \frac{\pi}{24} \).

Next, we need to find \( A^5 \). According to De Moivre's theorem, we can express \( A^n \) in terms of \( \cos(n\theta) \) and \( \sin(n\theta) \):

\[ A^n = \begin{pmatrix} \cos(n\theta) & i \sin(n\theta) \\ i \sin(n\theta) & \cos(n\theta) \end{pmatrix} \]

For \( n = 5 \):

\[ A^5 = \begin{pmatrix} \cos(5\theta) & i \sin(5\theta) \\ i \sin(5\theta) & \cos(5\theta) \end{pmatrix} \]

Now, substituting \( \theta = \frac{\pi}{24} \):

\[ 5\theta = 5 \cdot \frac{\pi}{24} = \frac{5\pi}{24} \]

Thus,

\[ A^5 = \begin{pmatrix} \cos\left(\frac{5\pi}{24}\right) & i \sin\left(\frac{5\pi}{24}\right) \\ i \sin\left(\frac{5\pi}{24}\right) & \cos\left(\frac{5\pi}{24}\right) \end{pmatrix} \]

Now we denote the elements of \( A^5 \) as follows:

\[ A^5 = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \]

where \( a = \cos\left(\frac{5\pi}{24}\right) \), \( b = i \sin\left(\frac{5\pi}{24}\right) \), \( c = i \sin\left(\frac{5\pi}{24}\right) \), and \( d = \cos\left(\frac{5\pi}{24}\right) \).

### Step 1: Analyze the options

1. **Option 1**: \( a^2 - d^2 = 0 \)
- Since \( a = d \), this option is true.

2. **Option 2**: \( a^2 - c^2 = 1 \)
- Here, \( c = i \sin\left(\frac{5\pi}{24}\right) \), thus:
\[ a^2 - c^2 = \cos^2\left(\frac{5\pi}{24}\right) - (i \sin\left(\frac{5\pi}{24}\right))^2 = \cos^2\left(\frac{5\pi}{24}\right) + \sin^2\left(\frac{5\pi}{24}\right) = 1 \]
- This option is true.

3. **Option 3**: \( a^2 - b^2 = \frac{1}{2} \)
- Here, \( b = i \sin\left(\frac{5\pi}{24}\right) \), thus:
\[ a^2 - b^2 = \cos^2\left(\frac{5\pi}{24}\right) - (i \sin\left(\frac{5\pi}{24}\right))^2 = \cos^2\left(\frac{5\pi}{24}\right) + \sin^2\left(\frac{5\pi}{24}\right) = 1 \]
- This option is false since it should equal 1, not \( \frac{1}{2} \).

4. **Option 4**: \( a + b^2 \) is between 0 and 1.
- We calculate:
\[ a + b^2 = \cos\left(\frac{5\pi}{24}\right) - \sin^2\left(\frac{5\pi}{24}\right) = \cos\left(\frac{5\pi}{24}\right) - (1 - \cos^2\left(\frac{5\pi}{24}\right)) \]
\[ = 2\cos^2\left(\frac{5\pi}{24}\right) - 1 \]
- This value is also between 0 and 1.

### Conclusion

The option that is **not true** is **Option 3**: \( a^2 - b^2 = \frac{1}{2} \).