Daily math problems for August 9, 2026 — Trigonometry, Calculus, Matrices & more

$\sin\dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}$ — a notable value from the unit circle.

$\bullet$ Simplify the limit by factoring numerator and denominator:

$$\lim_{k\to x}\frac{(k-4)(k+4)}{(k-4)(k-1)}=\frac{x+4}{x-1}$$

$\bullet$ The determinant has a row of zeros (with $\log_3 1=0$), so it contributes $0$:

$$f(x)=\frac{x+4}{x-1}$$

$\bullet$ Evaluate at $x=4$ and $x=0$:

$$f(4)=\tfrac{8}{3},\quad f(0)=-4$$

$\bullet$ Sum them:

$$f(4)+f(0)=\tfrac{8}{3}-4=-\tfrac{4}{3}$$

$f(0) = -1 < 0$ and $f(1) = 1 > 0$. Since $f$ is continuous, by the IVT there exists $c \in (0, 1)$ with $f(c) = 0$.

$\bullet$ Expand along the first row:

$$\det A = m(m^2-1) - 1\cdot(m-0) + 0$$

$\bullet$ Simplify:

$$\det A = m^3 - 2m = m(m^2-2)$$

$\bullet$ Set to zero:

$$m(m^2-2)=0$$

$\bullet$ Solutions:

$$m\in\{-\sqrt{2},\,0,\,\sqrt{2}\}$$

Using the closed forms $\cos 36^\circ = \tfrac{1+\sqrt{5}}{4}$ and $\cos 72^\circ = \tfrac{\sqrt{5}-1}{4}$, the difference is $\tfrac{(1+\sqrt{5}) - (\sqrt{5}-1)}{4} = \tfrac{2}{4} = \tfrac{1}{2}$.

$2\sin x \cos x - \sin x = 0 \Rightarrow \sin x (2\cos x - 1) = 0$. $\sin x = 0$: $x \in \{0, \pi\}$. $\cos x = 1/2$: $x \in \{\pi/3, 5\pi/3\}$. Total: $4$.

$A = 2000 \cdot 1.05^2 = 2000 \cdot 1.1025 = \$2{,}205$.

By Stirling, $\sqrt[n]{n!} \sim \dfrac{n}{e}\,(2\pi n)^{1/(2n)}$. The factor $(2\pi n)^{1/(2n)} \to 1$, so $\dfrac{\sqrt[n]{n!}}{n} \to \dfrac{1}{e}$.

$A = \begin{pmatrix} 2 & 3 \\ 1 & -1 \end{pmatrix}$ — coefficients of $x$ and $y$ in each equation.

$S_n = \dfrac{n(3n + 7)}{2}$. Compute: $S_{17} = \dfrac{17 \cdot 58}{2} = 493 < 500$ and $S_{18} = \dfrac{18 \cdot 61}{2} = 549 \ge 500$. Hence the smallest $n$ is $18$.