Daily math problems for August 12, 2026 — Trigonometry, Logaritmi, Calculus & more

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

Pentru a fi un logaritm valid trebuie ca baza $b>0,b\mathrm{\ne }1$ și argumentul $x>0$. Singura expresie care îndeplinește toate condițiile este ${\log }_{4}16$.

$\bullet$ Evaluate the constants: $\cos 100\pi =1$ and $\sqrt{{25}^3}=125$.

$\bullet$ Simplify the logarithm: ${\log }_2({32}^x)=x{\log }_{2}32=5x$, so $({\log }_2{32}^x{)}^3=125x^3$.

$\bullet$ The factor $125$ cancels and $x^3\mathrm{/}{x}^3=1$, leaving:

$$f(x)=\frac{125x^3}{125x^3\cdot x^{1\mathrm{/}4}}=x^{-1\mathrm{/}4}$$

$\bullet$ Differentiate by the power rule:

$$f^{\mathrm{\prime }}(x)=-\frac{1}{4}{x}^{-5\mathrm{/}4}$$

$\bullet$ Substitute $x=2$:

$$f^{\mathrm{\prime }}(2)=-\frac{1}{4\cdot 2^{5\mathrm{/}4}}$$

$\bullet$ Build the pieces for $A(1)=\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)$:

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

$$2A(1{)}^t=\left(\begin{array}{cc}2& 0\\ 4& 2\end{array}\right)$$

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

$\bullet$ Assemble $f(A(1))=A^2+2A^t-A^{-1}$ and take the determinant:

$$f(A(1))=\left(\begin{array}{cc}2& 6\\ 4& 2\end{array}\right),\det =4-24=-20$$

$\bullet$ For $A(-1)=\left(\begin{array}{cc}-1& 2\\ 0& 1\end{array}\right)$ we have $A(-1{)}^2=I$ and:

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

$\bullet$ Assemble $f(A(-1))$:

$$f(A(-1))=\left(\begin{array}{cc}0& -2\\ 4& 2\end{array}\right),\det =0-(-8)=8$$

$\bullet$ Multiply the two determinants:

$$\det \text{\hspace{0.17em}⁣}(f(A(1))\cdot f(A(-1)))=(-20)(8)=-160$$

$I_n+I_{n+1}={\int }_0^1{\displaystyle \frac{x^n+x^{n+1}}{1+x}}dx={\int }_0^1{x}^{n}dx={\displaystyle \frac{1}{n+1}}$.

${\displaystyle \frac{4\cdot 5\cdot 6}{3}}={\displaystyle \frac{120}{3}}=40$. (Direct: $2+6+12+20=40$.)

$A(x)=10x-x^2$. $A^{\mathrm{\prime }}(x)=10-2x=0\Rightarrow x=5$. The maximum-area rectangle is the square $5\times 5$, with area $25$.

$\int x{e}^{x}dx=x{e}^x-\int e^{x}dx=(x-1)e^x+C$. Evaluating from $0$ to $1$: $(0)\cdot e-(-1)\cdot 1=0-(-1)=1$.

The fifth roots of unity sum to $0$. Subtracting the root $1$ leaves a sum of $-1$, and taking real parts gives $\cos \frac{2\pi }{5}+\cos \frac{4\pi }{5}+\cos \frac{6\pi }{5}+\cos \frac{8\pi }{5}=-1$.

$P^{\mathrm{\prime }}(x)>0$ everywhere, so $P$ is strictly increasing on $\mathbb{R}$. Combined with ${\lim }_{\pm \mathrm{\infty }}P=\pm \mathrm{\infty }$, $P$ has exactly one real root by IVT.