Diagonalizing 2x2 Matrix and Finding Eigenvalues of 2x2 Matrix yields different results in Wolfram. Why?

Question

I want to find the eigenvalues of the matrix, $$ \begin{bmatrix} 0 & cos(k)+sin(k) \\ cos(k)+sin(k) & 0\\ \end{bmatrix}.$$

With diagonalize {{0, cos(k)+sin(k)}, {cos(k)+sin(k), 0}} I get the two eigenvalues: $$\pm\sqrt{sin(2k)+1}$$ With eigenvalues {{0, cos(k)+sin(k)}, {cos(k)+sin(k), 0}} I get: $$\pm(cos(k)+sin(k))$$ Why do I get different results?

Answer 1

$$\pm \sqrt{\sin(2k) + 1} = \pm \sqrt{(\sin(k) + \cos(k))^2} = \pm (\sin(k) + \cos(k))$$

Answer 2

Both answers are correct. For each real number k, the number $\sin(2k)+1$ is non-negative; therefore, it has $2$ square roots. Note that$$\bigl(\cos(k)+\sin(k)\bigr)^2=1+2\sin(k)\cos(k)=1+\sin(2k).$$Therefore, the square roots of $\sin(2k)+1$ are $\pm\bigl(\cos(k)+\sin(k)\bigr)$.