Daily math problems for August 29, 2026 — Trigonometric identities, Algebra, Calculus & more

By the addition formula, the expression equals $\sin ({75}^{\circ }+{15}^{\circ })=\sin {90}^{\circ }=1$.

If $1+i$ is a root, so is its conjugate $1-i$. So the polynomial must have at least these two roots — minimum degree $2$. (E.g., $(X-1-i)(X-1+i)=X^2-2X+2$.)

$V=\pi {\int }_0^3{x}^{2}dx=\pi \cdot {\displaystyle \frac{27}{3}}=9\pi$. (Equivalently, $V_{\text{cone}}={\displaystyle \frac{1}{3}}\pi r^{2}h={\displaystyle \frac{1}{3}}\pi \cdot 9\cdot 3=9\pi$.)

Invertibility ⇔ non-zero determinant.

$V=\pi {\int }_0^1(e^x{)}^{2}dx=\pi {\int }_0^1{e}^{2x}dx=\pi \cdot {\displaystyle \frac{e^{2x}}{2}}{\mid }_0^1={\displaystyle \frac{\pi }{2}}(e^2-1)$.

$\vec{v}=2\vec{u}$, so the two vectors are collinear (parallel).

$\bullet$ Simplify exponent and argument:

$${\log }_{2}16=4,i^4=1$$

$\bullet$ The function reduces to:

$$f(x)=x^4-\cos (\pi x)$$

$\bullet$ Differentiate:

$$f^{\mathrm{\prime }}(x)=4x^3+\pi \sin (\pi x)$$

$\bullet$ Evaluate at $x=1$, using $\sin \pi =0$:

$$f^{\mathrm{\prime }}(1)=4+\pi \cdot 0=4$$

$\bullet$ Evaluate each of the nine limits to fill the matrix:

$$\left(\begin{array}{ccc}\sqrt{3}\mathrm{/}2& 2& 1\\ 3& 4& 4\\ 0& 3& 1\end{array}\right)$$

$\bullet$ Expand along row 1. The cofactor of the $(1,1)$ entry is $4\cdot 1-4\cdot 3=-8$:

$$\frac{\sqrt{3}}{2}\cdot (-8)=-4\sqrt{3}$$

$\bullet$ Cofactors of the $(1,2)$ and $(1,3)$ entries are $3\cdot 1-4\cdot 0=3$ and $3\cdot 3-4\cdot 0=9$:

$$-2\cdot 3+1\cdot 9=3$$

$\bullet$ Add the three contributions:

$$\det =3-4\sqrt{3}$$

${\displaystyle \frac{p}{q}}$ runs over $\text{\hspace{0.17em}⁣}\{\pm 1,\pm {\displaystyle \frac{1}{2}}\}$.

${\displaystyle \frac{2x^2+3x+1}{x-1}}=2x+5+{\displaystyle \frac{6}{x-1}}$. As $x\to \pm \mathrm{\infty }$, the remainder vanishes, leaving the slant asymptote $y=2x+5$.