Let $$\theta = \frac{\pi}{5}$$ and $$A = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$$. If $$B = A + A^4$$, then $$\det(B)$$:

We have the angle $$\theta=\dfrac{\pi}{5}\;(=36^{\circ})$$ and the matrix

$$ A=\begin{bmatrix} \cos\theta & \sin\theta\\ -\sin\theta& \cos\theta \end{bmatrix}. $$

This is the standard rotation matrix, so $$\det(A)=1.$$ A power of a rotation matrix is again a rotation matrix: in fact

$$ A^{n}=\begin{bmatrix} \cos(n\theta) & \sin(n\theta)\\ -\sin(n\theta)& \cos(n\theta) \end{bmatrix} \quad\text{for every integer }n. $$

Taking $$n=4$$ we obtain

$$ A^{4}=\begin{bmatrix} \cos(4\theta) & \sin(4\theta)\\ -\sin(4\theta)& \cos(4\theta) \end{bmatrix}. $$

Now define $$B=A+A^{4}.$$ Adding the two rotation matrices entry-wise we get

$$ B=\begin{bmatrix} \cos\theta+\cos4\theta & \;\sin\theta+\sin4\theta\\ -(\sin\theta+\sin4\theta) & \;\cos\theta+\cos4\theta \end{bmatrix}. $$

For any matrix of the form $$\begin{bmatrix}a & b\\ -b & a\end{bmatrix},$$ the determinant is $$a^{2}+b^{2},$$ because

$$ \det\begin{bmatrix}a & b\\ -b & a\end{bmatrix}=a\cdot a-(-b)\,b=a^{2}+b^{2}. $$

So, with $$a=\cos\theta+\cos4\theta,\qquad b=\sin\theta+\sin4\theta,$$ we have

$$ \det(B)=\bigl(\cos\theta+\cos4\theta\bigr)^{2}+\bigl(\sin\theta+\sin4\theta\bigr)^{2}. $$

We evaluate the two trigonometric sums. First use the sum-to-product identity

$$ \cos x+\cos y=2\cos\!\left(\dfrac{x+y}{2}\right)\cos\!\left(\dfrac{x-y}{2}\right). $$

Putting $$x=\theta,\;y=4\theta$$ gives

$$ \cos\theta+\cos4\theta =2\cos\!\left(\dfrac{\theta+4\theta}{2}\right)\cos\!\left(\dfrac{\theta-4\theta}{2}\right) =2\cos\!\left(\dfrac{5\theta}{2}\right)\cos\!\left(-\dfrac{3\theta}{2}\right). $$

Because $$5\theta=\pi,$$ we have $$\dfrac{5\theta}{2}=\dfrac{\pi}{2},$$ and $$\cos\dfrac{\pi}{2}=0.$$ Hence

$$ \cos\theta+\cos4\theta=0. $$

Next use the sum-to-product identity for sines,

$$ \sin x+\sin y=2\sin\!\left(\dfrac{x+y}{2}\right)\cos\!\left(\dfrac{x-y}{2}\right), $$

again with $$x=\theta,\;y=4\theta:$$

$$ \sin\theta+\sin4\theta =2\sin\!\left(\dfrac{5\theta}{2}\right)\cos\!\left(-\dfrac{3\theta}{2}\right) =2\sin\!\left(\dfrac{\pi}{2}\right)\cos\!\left(-\dfrac{3\theta}{2}\right). $$

Since $$\sin\dfrac{\pi}{2}=1$$ and $$\cos(-x)=\cos x,$$ we get

$$ \sin\theta+\sin4\theta =2\cos\!\left(\dfrac{3\theta}{2}\right) =2\cos\!\left(\dfrac{3\pi}{10}\right). $$

Hence

$$ \det(B)=0^{2}+\Bigl(2\cos\dfrac{3\pi}{10}\Bigr)^{2} =4\cos^{2}\dfrac{3\pi}{10}. $$

The exact value of $$\cos\dfrac{3\pi}{10}=\cos54^{\circ}$$ is known:

$$ \cos54^{\circ}=\dfrac{\sqrt{10-2\sqrt5}}{4}. $$

Squaring and multiplying by $$4$$ we obtain

$$ \det(B)=4\left(\dfrac{10-2\sqrt5}{16}\right) =\dfrac{10-2\sqrt5}{4} =\dfrac{5-\sqrt5}{2}. $$

The numerical value is

$$ \dfrac{5-\sqrt5}{2}\approx\dfrac{5-2.236}{2}\approx\dfrac{2.764}{2}\approx1.382. $$

This lies strictly between $$1$$ and $$2.$$ Therefore $$\det(B)$$ belongs to the interval $$(1,2).$$

Hence, the correct answer is Option D.