Daily math problems for October 3, 2026 — Trigonometry, Trigonometric identities, Limits & more

A standard unit-circle value: $\sin {\displaystyle \frac{\pi }{6}}={\displaystyle \frac{1}{2}}$.

$\sin {75}^{\circ }+\sin {15}^{\circ }=2\sin {\displaystyle \frac{{90}^{\circ }}{2}}\cos {\displaystyle \frac{{60}^{\circ }}{2}}=2\sin {45}^{\circ }\cos {30}^{\circ }=2\cdot {\displaystyle \frac{\sqrt{2}}{2}}\cdot {\displaystyle \frac{\sqrt{3}}{2}}={\displaystyle \frac{\sqrt{6}}{2}}$.

$(\sin x+\cos x{)}^2=1+2\sin x\cos x=1+\sin 2x=\frac{1}{2}$, so $\sin 2x=-\frac{1}{2}$.

$\bullet$ Set $L={\lim }_{x\to \mathrm{\infty }}(1+x{)}^{1\mathrm{/}x}$ and take the logarithm:

$$\ln L=\underset{x\to \mathrm{\infty }}{\lim }\frac{\ln (1+x)}{x}$$

$\bullet$ Logarithm grows slower than $x$, so the ratio vanishes:

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\ln (1+x)}{x}=0$$

$\bullet$ Exponentiate:

$$L=e^0=1$$

$f^{\mathrm{\prime }}(x)=3x^2-3$, $f^{\mathrm{\prime }\mathrm{\prime }}(x)=6x$. $f^{\mathrm{\prime }\mathrm{\prime }}$ changes sign at $x=0$, so the inflection is at $x=0$.

$F(x)=\sin x$. $F(\pi \mathrm{/}2)-F(0)=1-0=1$.

$\overrightarrow{AB}=(3,5)$, $\overrightarrow{AC}=(-1,3)$. $\det =3\cdot 3-5\cdot (-1)=14$. Area $={\displaystyle \frac{14}{2}}=7$.

$u=x$, $du=dx$, $v=\sin x$. $\int x\cos xdx=x\sin x-\int \sin xdx=x\sin x+\cos x+C$.

$\text{\hspace{0.17em}⁣}{(1+{\displaystyle \frac{0.12}{12}})}^{12}-1={1.01}^{12}-1\approx 1.1268-1=0.1268\approx 12.68\mathrm{\%}$.

$x^2>9\iff \mathrm{\mid }x\mathrm{\mid }>3\iff x<-3$ or $x>3$. Solution: $(-\mathrm{\infty },-3)\cup (3,+\mathrm{\infty })$.