Daily math problems for October 7, 2026 — Trigonometry, Logic, Vectors & more

Reference angle $\frac{\pi }{4}$. In $[0,2\pi )$ cosine is $\frac{\sqrt{2}}{2}$ at $x=\frac{\pi }{4}$ (Q1) and $x=\frac{7\pi }{4}$ (Q4).

$1+3+5+\cdots +17=9^2=81$. (Direct check: $1+3+5+7+9+11+13+15+17=81$.)

$\overrightarrow{AB}=(4-1)\vec{i}+(6-2)\vec{j}=3\vec{i}+4\vec{j}$.

$R(\theta )R(-\theta )=I_2$, so $R(\theta {)}^{-1}=R(-\theta )$.

Pair each subset $S$ with $S\mathrm{△}\{1\}$ (i.e., toggle the element $1$). This pairing has no fixed points and flips the parity of the sum, so even-sum and odd-sum subsets are equinumerous: $2^{10}\mathrm{/}2=512$.

$2{\cos }^{2}x+\cos x-1=0\Rightarrow (2\cos x-1)(\cos x+1)=0$. $\cos x=1\mathrm{/}2$: 2 solutions ($\pi \mathrm{/}3,5\pi \mathrm{/}3$). $\cos x=-1$: 1 solution ($\pi$). Total: $3$.

$\mathrm{\mid }A\times B\mathrm{\mid }=3\cdot 2=6$. The product is $\{(1,x),(1,y),(2,x),(2,y),(3,x),(3,y)\}$.

$7^2=49$, $7^4={49}^2=2401\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}100)$. Since $2024=4\cdot 506$, we have $7^{2024}=(7^4{)}^{506}\equiv 1^{506}=1(\mathrm{m}\mathrm{o}\mathrm{d}100)$. Last two digits: $01$.

$n\ln \text{\hspace{0.17em}⁣}(1+\frac{1}{n^2})\sim n\cdot \frac{1}{n^2}=\frac{1}{n}\to 0$, so the limit equals $e^0=1$.

$\overrightarrow{AB}=(4,0)$ and $\overrightarrow{DC}=(4,0)$ — same vector, so $ABCD$ is a parallelogram. Base $=4$, height $=3$. Area $=12$.