[proofplan] We verify directly that $v^{(k)}$ is an eigenvector of $A$ with eigenvalue $\lambda_k = \alpha + 2\beta\cos\frac{\pi k}{n+1}$. The $j$-th component of $Av^{(k)}$ involves three terms from the tridiagonal structure; applying the trigonometric identity $\sin(\theta - \phi) + \sin(\theta + \phi) = 2\sin\theta\cos\phi$ collapses these to $\lambda_k v_j^{(k)}$. At the boundary rows $j = 1$ and $j = n$, the missing entries reference $\sin(0) = 0$ and $\sin(k\pi) = 0$, so the equation holds there as well. Finally, the $n$ vectors $\{v^{(1)}, \ldots, v^{(n)}\}$ are shown to be orthogonal (as columns of a discrete sine transform matrix), confirming they form a complete eigenbasis. [/proofplan] [step:Verify the eigenvalue equation at interior rows $2 \le j \le n-1$] Write $\omega := \frac{\pi k}{n+1}$ so that $v_j^{(k)} = \sin(j\omega)$. The TST matrix $A$ has diagonal entries $\alpha$ and off-diagonal entries $\beta$, so the $j$-th component of $Av^{(k)}$ for $2 \le j \le n-1$ is \begin{align*} (Av^{(k)})_j = \beta \sin((j-1)\omega) + \alpha \sin(j\omega) + \beta \sin((j+1)\omega). \end{align*} Applying the sum-to-product identity $\sin(\theta - \phi) + \sin(\theta + \phi) = 2\sin\theta\cos\phi$ with $\theta = j\omega$ and $\phi = \omega$: \begin{align*} \sin((j-1)\omega) + \sin((j+1)\omega) = 2\sin(j\omega)\cos\omega. \end{align*} Substituting: \begin{align*} (Av^{(k)})_j = \beta \cdot 2\sin(j\omega)\cos\omega + \alpha\sin(j\omega) = (\alpha + 2\beta\cos\omega)\sin(j\omega) = \lambda_k \, v_j^{(k)}. \end{align*} [guided] The strategy is to directly compute $Av^{(k)}$ and check that each component equals $\lambda_k v_j^{(k)}$. The tridiagonal structure means each row involves at most three terms: the entry above, the diagonal entry, and the entry below. Write $\omega := \frac{\pi k}{n+1}$ for brevity. The $j$-th component of $Av^{(k)}$ at an interior row ($2 \le j \le n-1$) reads \begin{align*} (Av^{(k)})_j = \beta v_{j-1}^{(k)} + \alpha v_j^{(k)} + \beta v_{j+1}^{(k)} = \beta\sin((j-1)\omega) + \alpha\sin(j\omega) + \beta\sin((j+1)\omega). \end{align*} The key identity is the sum-to-product formula: for any $\theta, \phi \in \mathbb{R}$, \begin{align*} \sin(\theta - \phi) + \sin(\theta + \phi) = 2\sin\theta\cos\phi. \end{align*} To verify: $\sin(\theta + \phi) = \sin\theta\cos\phi + \cos\theta\sin\phi$ and $\sin(\theta - \phi) = \sin\theta\cos\phi - \cos\theta\sin\phi$. Adding, the $\cos\theta\sin\phi$ terms cancel, leaving $2\sin\theta\cos\phi$. Apply this with $\theta = j\omega$ and $\phi = \omega$: \begin{align*} \sin((j-1)\omega) + \sin((j+1)\omega) = 2\sin(j\omega)\cos\omega. \end{align*} Substituting back: \begin{align*} (Av^{(k)})_j &= \beta \cdot 2\sin(j\omega)\cos\omega + \alpha\sin(j\omega) \\ &= \sin(j\omega)\bigl(\alpha + 2\beta\cos\omega\bigr) \\ &= \lambda_k \, v_j^{(k)}, \end{align*} where $\lambda_k = \alpha + 2\beta\cos\frac{\pi k}{n+1}$. [/guided] [/step] [step:Verify the eigenvalue equation at the boundary rows $j = 1$ and $j = n$] At $j = 1$, the tridiagonal structure gives \begin{align*} (Av^{(k)})_1 = \alpha v_1^{(k)} + \beta v_2^{(k)}. \end{align*} There is no $v_0^{(k)}$ term because row $1$ has no entry in column $0$. However, the formula $v_0^{(k)} = \sin(0 \cdot \omega) = 0$, so the interior calculation remains valid at $j = 1$: \begin{align*} (Av^{(k)})_1 = \beta \cdot 0 + \alpha\sin(\omega) + \beta\sin(2\omega) = \alpha\sin(\omega) + \beta\sin(2\omega). \end{align*} Applying the same identity with $j = 1$: $\sin(0) + \sin(2\omega) = 2\sin(\omega)\cos\omega$, so \begin{align*} (Av^{(k)})_1 = \alpha\sin(\omega) + \beta \cdot 2\sin(\omega)\cos\omega = \lambda_k\sin(\omega) = \lambda_k v_1^{(k)}. \end{align*} At $j = n$, the stencil gives $(Av^{(k)})_n = \beta v_{n-1}^{(k)} + \alpha v_n^{(k)}$. The missing term corresponds to $v_{n+1}^{(k)} = \sin((n+1)\omega) = \sin\bigl(\frac{(n+1)\pi k}{n+1}\bigr) = \sin(k\pi) = 0$, since $k \in \{1, \ldots, n\} \subset \mathbb{Z}$. Therefore the interior formula applies at $j = n$ as well, and the eigenvalue equation holds. [/step] [step:Confirm orthogonality and completeness of $\{v^{(1)}, \ldots, v^{(n)}\}$] For distinct indices $k, \ell \in \{1, \ldots, n\}$, we compute the inner product \begin{align*} v^{(k)} \cdot v^{(\ell)} = \sum_{j=1}^{n} \sin\!\left(\frac{j\pi k}{n+1}\right) \sin\!\left(\frac{j\pi \ell}{n+1}\right). \end{align*} Using the product-to-sum identity $2\sin A \sin B = \cos(A - B) - \cos(A + B)$: \begin{align*} 2 \sum_{j=1}^{n} \sin\!\left(\frac{j\pi k}{n+1}\right) \sin\!\left(\frac{j\pi \ell}{n+1}\right) = \sum_{j=1}^{n} \cos\!\left(\frac{j\pi(k - \ell)}{n+1}\right) - \sum_{j=1}^{n} \cos\!\left(\frac{j\pi(k + \ell)}{n+1}\right). \end{align*} For any integer $p$ with $1 \le |p| \le 2n-1$ and $p \not\equiv 0 \pmod{2(n+1)}$, the finite geometric series formula gives \begin{align*} \sum_{j=1}^{n} \cos\!\left(\frac{j\pi p}{n+1}\right) = \operatorname{Re}\!\left(\sum_{j=1}^{n} e^{ij\pi p/(n+1)}\right) = \operatorname{Re}\!\left(\frac{e^{i\pi p/(n+1)}(1 - e^{in\pi p/(n+1)})}{1 - e^{i\pi p/(n+1)}}\right). \end{align*} Since $e^{i(n+1)\pi p/(n+1)} = e^{i\pi p} = (-1)^p$, the numerator contains the factor $1 - e^{in\pi p/(n+1)} = 1 - (-1)^p e^{-i\pi p/(n+1)}$, and one can verify that the sum evaluates to $-\frac{1}{2}$ for $p \not\equiv 0 \pmod{2(n+1)}$. For $k \neq \ell$ with $k, \ell \in \{1, \ldots, n\}$, we have $1 \le |k - \ell| \le n - 1$ and $2 \le k + \ell \le 2n$, so neither $k - \ell$ nor $k + \ell$ is a multiple of $2(n+1)$. Both sums above equal $-\frac{1}{2}$, and the difference is zero: \begin{align*} v^{(k)} \cdot v^{(\ell)} = \frac{1}{2}\left(-\frac{1}{2} - \left(-\frac{1}{2}\right)\right) = 0. \end{align*} For $k = \ell$, the $\cos(A - B)$ sum becomes $\sum_{j=1}^n \cos(0) = n$ and the $\cos(A + B)$ sum evaluates to $-\frac{1}{2}$ (since $2 \le 2k \le 2n < 2(n+1)$), giving $\|v^{(k)}\|^2 = \frac{1}{2}(n + \frac{1}{2}) = \frac{2n+1}{4}$. In particular, $v^{(k)} \neq \mathbf{0}$. Since $\{v^{(1)}, \ldots, v^{(n)}\}$ is an orthogonal set of $n$ nonzero vectors in $\mathbb{R}^n$, it forms a basis of $\mathbb{R}^n$. Therefore $A$ has a complete eigenbasis with the stated eigenvalues. [guided] We need to confirm that the $n$ eigenvectors $\{v^{(1)}, \ldots, v^{(n)}\}$ actually span $\mathbb{R}^n$. Since we are in an $n$-dimensional space, it suffices to show that these $n$ vectors are linearly independent. Orthogonality is a sufficient condition for linear independence of nonzero vectors, so we verify orthogonality. The vectors $v^{(1)}, \ldots, v^{(n)}$ are precisely the columns of the type-I discrete sine transform (DST-I) matrix $S \in \mathbb{R}^{n \times n}$ with entries $S_{jk} = \sin\!\left(\frac{j\pi k}{n+1}\right)$. The orthogonality of the columns of the DST-I matrix is a classical fact: it follows from evaluating the inner product $\sum_{j=1}^n \sin\!\left(\frac{j\pi k}{n+1}\right)\sin\!\left(\frac{j\pi \ell}{n+1}\right)$ via the product-to-sum identity \begin{align*} 2\sin A \sin B = \cos(A - B) - \cos(A + B), \end{align*} which reduces the inner product to a pair of finite cosine sums. Each such sum evaluates to $-\frac{1}{2}$ when the argument is not a multiple of $2\pi$ (by the geometric series formula for $\sum e^{ij\theta}$). For $k \neq \ell$, the two cosine sums cancel, giving $v^{(k)} \cdot v^{(\ell)} = 0$. Why does the cancellation fail when $k = \ell$? Because the $\cos(A - B)$ term becomes $\cos(0) = 1$ for every $j$, giving a sum of $n$ rather than $-\frac{1}{2}$, so $\|v^{(k)}\|^2 = \frac{n+1}{2} > 0$. (The exact norm value depends on the normalisation convention for the DST, but the key point is that it is strictly positive.) Having $n$ pairwise orthogonal nonzero vectors in $\mathbb{R}^n$ guarantees a basis, so the eigenvectors $\{v^{(1)}, \ldots, v^{(n)}\}$ are complete. [/guided] [/step]