Daily math problems for August 7, 2026 — Algebra, Calculus, Trigonometry & more

A ring: abelian group under $+$, associative $\cdot$, and distributivity of $\cdot$ over $+$. (Some definitions also require a multiplicative identity.)

$I_0={\int }_0^{1}1dx=1$.

For polynomials, substitute directly: $4+6-1=9$.

$\bullet$ Recognise a binomial square:

$${\sin }^{4}7\mathrm{°}+2{\sin }^{2}7\mathrm{°}{\cos }^{2}7\mathrm{°}+{\cos }^{4}7\mathrm{°}=({\sin }^{2}7\mathrm{°}+{\cos }^{2}7\mathrm{°}{)}^2$$

$\bullet$ Pythagoras kills the angle:

$$({\sin }^{2}7\mathrm{°}+{\cos }^{2}7\mathrm{°}{)}^2=1$$

$\bullet$ Evaluate the limit:

$$\underset{x\to 0}{\lim }{\displaystyle \frac{\sin 5x}{x}}=5$$

$\bullet$ Decode the log:

$${\log }_{\sqrt{7}}49={\displaystyle \frac{2}{1\mathrm{/}2}}=4$$

$\bullet$ Combine:

$$1+5-4=2$$

Lagrange: in a finite group, the order of any subgroup (and hence the order of any element, which is the order of the cyclic subgroup it generates) divides $\mathrm{\mid }G\mathrm{\mid }$.

$5$ is prime, so ${\mathbb{Z}}_5$ is a field. (Every non-zero element has a multiplicative inverse.)

$\bullet$ Evaluate every entry:

$$\cos 60\mathrm{°}=\sin 30\mathrm{°}=\frac{1}{2},\tan 45\mathrm{°}=1,{\log }_{2}4=2$$

$\bullet$ The matrix becomes:

$$A=\left(\begin{array}{cc}1\mathrm{/}2& 1\\ 2& 1\mathrm{/}2\end{array}\right)$$

$\bullet$ Compute the determinant:

$$\det A=\frac{1}{4}-2=-\frac{7}{4}$$

$\bullet$ Apply $\det (A^{-1})=1\mathrm{/}\det A$:

$$\det (A^{-1})=-\frac{4}{7}$$

After both moves, the value is $0.8\cdot 1.2=0.96$ — a $4\mathrm{\%}$ net loss.

${\displaystyle \frac{1}{n}}\sum (x_i+5)={\displaystyle \frac{1}{n}}\sum x_i+5=\bar{x}+5$.

Folosind schimbarea de bază, ${\log }_k(k+1)={\displaystyle \frac{\ln (k+1)}{\ln k}}$. Produsul telescopează: ${\displaystyle \frac{\ln 3}{\ln 2}}\cdot {\displaystyle \frac{\ln 4}{\ln 3}}\cdots {\displaystyle \frac{\ln 256}{\ln 255}}={\displaystyle \frac{\ln 256}{\ln 2}}={\log }_{2}256=8$.