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

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

Pentru a fi un logaritm valid trebuie ca baza $b > 0,\ b \neq 1$ 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 $x^3/x^3=1$, leaving:

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

$\bullet$ Differentiate by the power rule:

$$f'(x)=-\tfrac{1}{4}x^{-5/4}$$

$\bullet$ Substitute $x=2$:

$$f'(2)=-\tfrac{1}{4\cdot 2^{5/4}}$$

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

$$A(1)^2=\begin{pmatrix}1&4\\0&1\end{pmatrix}$$

$$2A(1)^t=\begin{pmatrix}2&0\\4&2\end{pmatrix}$$

$$A(1)^{-1}=\begin{pmatrix}1&-2\\0&1\end{pmatrix}$$

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

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

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

$$2A(-1)^t=\begin{pmatrix}-2&0\\4&2\end{pmatrix},\ A(-1)^{-1}=\begin{pmatrix}-1&2\\0&1\end{pmatrix}$$

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

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

$\bullet$ Multiply the two determinants:

$$\det\!\bigl(f(A(1))\cdot f(A(-1))\bigr)=(-20)(8)=-160$$

$I_n + I_{n+1} = \int_0^1 \dfrac{x^n + x^{n+1}}{1 + x} \, dx = \int_0^1 x^n \, dx = \dfrac{1}{n + 1}$.

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

$A(x) = 10x - x^2$. $A'(x) = 10 - 2x = 0 \Rightarrow x = 5$. 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 $\cos\tfrac{2\pi}{5}+\cos\tfrac{4\pi}{5}+\cos\tfrac{6\pi}{5}+\cos\tfrac{8\pi}{5}=-1$.

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