Daily math problems for August 17, 2026 — Trigonometry, Algebra, Probability & more

$x=\pi \mathrm{/}6$ and $x=\pi -\pi \mathrm{/}6=5\pi \mathrm{/}6$. Two solutions.

$3\cdot 5=15=2\cdot 7+1$, so $\widehat{3}\cdot \widehat{5}=\widehat{1}$.

A fair coin has two equally likely outcomes; the probability of heads is ${\displaystyle \frac{1}{2}}$.

$(1+i{)}^2=2i$, so $(1+i{)}^{10}=(2i{)}^5=32\cdot i^5=32i$.

Matrices form a ring (with $0$ matrix and identity matrix), but multiplication is non-commutative for $n\ge 2$. It's a non-commutative ring with unity.

$a=3$, $b=4$, so asymptotes: $y=\pm {\displaystyle \frac{4}{3}}x$.

The graph is a semicircle; rotating about the $x$-axis sweeps out a sphere of radius $R$. (Volume $={\displaystyle \frac{4}{3}}\pi R^3$.)

Midpoint: $(2,0)$. Segment $AB$ is horizontal, so the bisector is vertical through $(2,0)$: $x=2$.

$\bullet$ Square the first matrix using $i^2=-1$:

$${\left(\begin{array}{cc}3& 2\\ 0& -1\end{array}\right)}^2=\left(\begin{array}{cc}9& 4\\ 0& 1\end{array}\right)$$

$\bullet$ Evaluate the entries of the second matrix: $\lim x^2\mathrm{/}x!=0$, ${\log }_{3}81=4$, $i^{2022}=-1$, $\tan 225\mathrm{°}=1$:

$$\left(\begin{array}{cc}0& 4\\ -1& 1\end{array}\right)$$

$\bullet$ Add and take the determinant:

$$\left(\begin{array}{cc}9& 8\\ -1& 2\end{array}\right),\det =18-(-8)=26$$

$2\sin x\cos x-\sqrt{3}\cos x=0\Rightarrow \cos x(2\sin x-\sqrt{3})=0$.

$\bullet$ $\cos x=0\Rightarrow x\in \text{\hspace{0.17em}⁣}\{{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{3\pi }{2}}\}$

$\bullet$ $\sin x={\displaystyle \frac{\sqrt{3}}{2}}\Rightarrow x\in \text{\hspace{0.17em}⁣}\{{\displaystyle \frac{\pi }{3}},{\displaystyle \frac{2\pi }{3}}\}$

Total: $4$ solutions on $[0,2\pi )$.