Daily math problems for September 11, 2026 — Trigonometry, Algebraic identities, Combinatorics & more

$\sin x=\frac{\sqrt{3}}{2}$ at the reference angle $\frac{\pi }{3}$. Sine is positive in Q1 and Q2: $x=\frac{\pi }{3}$ and $x=\frac{2\pi }{3}$.

$(x+y{)}^2=(x^2+y^2)+2xy=13+12=25$, so $\mathrm{\mid }x+y\mathrm{\mid }=5$.

${\displaystyle \frac{n(n-1)}{2}}=15\Rightarrow n^2-n-30=0\Rightarrow n\in \{6,-5\}$. Only $n=6$ is admissible.

Using $1-\cos x=2{\sin }^2(x\mathrm{/}2)$: ${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{2{\sin }^2(x\mathrm{/}2)}{x^2}}={\displaystyle \frac{1}{2}}\underset{x\to 0}{\lim }{\left({\displaystyle \frac{\sin (x\mathrm{/}2)}{x\mathrm{/}2}}\right)}^2={\displaystyle \frac{1}{2}}\cdot 1={\displaystyle \frac{1}{2}}}$.

$\vec{u}\cdot \vec{v}=a+(-2)=a-2=0\Rightarrow a=2$.

$\mathrm{sgn}((ij))=(-1{)}^{2-1}=-1$.

For $z_1=2+b_{1}i$ and $z_2=2+b_{2}i$: $\overline{z_j}=2-b_{j}i$. Compute: $z_1\circ z_2=(2+b_{1}i)+(2+b_{2}i)-{\displaystyle \frac{1}{2}}(2-b_{1}i)-{\displaystyle \frac{1}{2}}(2-b_{2}i)=4+(b_1+b_2)i-2+{\displaystyle \frac{b_1+b_2}{2}}i=2+{\displaystyle \frac{3(b_1+b_2)}{2}}i\in H$. So $H$ is stable.

$\bullet$ Simplify each logarithm:

$${\log }_{\sqrt{2}}\text{\hspace{0.17em}⁣}\left(\frac{1}{x-3}\right)=-2{\log }_2(x-3)$$

$${\log }_{100}({10}^x)=\frac{x}{2}$$

$\bullet$ Combine the numerator:

$${\log }_2(x+25)-2{\log }_2(x-3)={\log }_2\frac{x+25}{(x-3{)}^2}$$

$\bullet$ The equation becomes:

$$\frac{{\log }_2\frac{x+25}{(x-3{)}^2}}{x\mathrm{/}2}=\frac{2}{7}$$

$${\log }_2\frac{x+25}{(x-3{)}^2}=\frac{x}{7}$$

$\bullet$ Test $x=7$:

$${\log }_2\frac{32}{16}={\log }_{2}2=1=\frac{7}{7}$$

${10}^3-10=990$, and $990\mathrm{/}6=165$. The induction proof factors $n^3-n=(n-1)n(n+1)$, which is the product of three consecutive integers and is therefore divisible by both $2$ and $3$, hence by $6$.

$\bullet$ The astroid meets the axes at $(\pm a,0)$ and $(0,\pm a)$, so it fits in a square of side $2a$.

$\bullet$ Given side $=4$:

$$2a=4\Rightarrow a=2$$

$\bullet$ Apply the astroid area formula:

$$\mathcal{A}=\frac{3\pi a^2}{8}=\frac{3\pi \cdot 4}{8}=\frac{3\pi }{2}$$