Daily math problems for September 2, 2026 — Logarithms, Trigonometry, Descriptive Statistics & Sampling & more

${\log }_{3}81=4$, ${\log }_{3}9=2$, so the ratio is $\frac{4}{2}=2$.

$\bullet$ Apply the difference-to-product identity:

$$\sin A-\sin B=2\cos \frac{A+B}{2}\sin \frac{A-B}{2}$$

$\bullet$ With $A=75\mathrm{°}$, $B=15\mathrm{°}$ the half-sum is $45\mathrm{°}$ and half-difference is $30\mathrm{°}$:

$$\sin 75\mathrm{°}-\sin 15\mathrm{°}=2\cos 45\mathrm{°}\sin 30\mathrm{°}$$

$\bullet$ Evaluate:

$$2\cdot \frac{\sqrt{2}}{2}\cdot \frac{1}{2}=\frac{\sqrt{2}}{2}$$

Sum $=5\cdot 7+8\cdot 8+7\cdot 9=35+64+63=162$. Mean $={\displaystyle \frac{162}{20}}=8.1$.

$d=\mathrm{\mid }-6\mathrm{\mid }\mathrm{/}\sqrt{4+4+1}=6\mathrm{/}3=2$.

$\sin x\ge 0$ on $[0,\pi \mathrm{/}2]$, so ${\sin }^{n}x\ge 0$ and the integral is $\ge 0$ (in fact $>0$ for finite $n$).

$\ker \phi ={\phi }^{-1}(\{e_H\})=\{a\in G\mid \phi (a)=e_H\}$.

Geometric series with $a=1$, $r=\frac{1}{2}$. Sum $=\frac{a}{1-r}=\frac{1}{1-1\mathrm{/}2}=2$.

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

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

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