Daily math problems for September 19, 2026 — Calculus, Complex Numbers, Trigonometry & more

$V=\pi {\int }_0^4(\sqrt{x}{)}^{2}dx=\pi {\int }_0^{4}xdx=\pi \cdot {\displaystyle \frac{16}{2}}=8\pi$.

$\bullet$ Compute the modulus of the base:

$$\mathrm{\mid }1+i\sqrt{3}\mathrm{\mid }=\sqrt{1+3}=2$$

$\bullet$ Modulus distributes over powers:

$$\mathrm{\mid }(1+i\sqrt{3}{)}^{2024}\mathrm{\mid }=2^{2024}$$

$\bullet$ The denominator is real and positive, so:

$$\mid {\displaystyle \frac{(1+i\sqrt{3}{)}^{2024}}{2^{2023}}}\mid ={\displaystyle \frac{2^{2024}}{2^{2023}}}=2$$

$\bullet$ Cosine is even, so $\cos (y-x)=\cos (x-y)$:

$$2\cos (x-y)=\sqrt{3}\Rightarrow \cos (x-y)=\frac{\sqrt{3}}{2}$$

$\bullet$ So $x-y=\frac{\pi }{6}$ (principal value).

$\bullet$ Substitute into the first equation:

$$\sin (x+y)+\sin \frac{\pi }{6}=\frac{3}{2}$$

$$\sin (x+y)=1\Rightarrow x+y=\frac{\pi }{2}$$

$\bullet$ Compute the tangent:

$$\tan \frac{x+y}{2}=\tan \frac{\pi }{4}=1$$

$n+1=11\Rightarrow n=10$.

$\sin x+\cos x=\sqrt{2}\text{\hspace{0.17em}⁣}(\sin x\cdot {\displaystyle \frac{\sqrt{2}}{2}}+\cos x\cdot {\displaystyle \frac{\sqrt{2}}{2}})=\sqrt{2}\sin \text{\hspace{0.17em}⁣}(x+{\displaystyle \frac{\pi }{4}})$.

Since $\sin$ has maximum $1$, the maximum of $f$ is $\sqrt{2}$, attained at $x={\displaystyle \frac{\pi }{4}}$.

The sum telescopes: ${\sum }_{k=1}^{99}({\displaystyle \frac{1}{k}}-{\displaystyle \frac{1}{k+1}})=1-{\displaystyle \frac{1}{100}}={\displaystyle \frac{99}{100}}$.

We need $a_{12}=a_{21}$, i.e. $a=3$.

In $V_4$, every non-identity element has order $2$. (This distinguishes it from ${\mathbb{Z}}_4$, which has an element of order $4$.)

$V=\pi {\int }_0^2(4x^2+4x+1)dx=\pi \text{\hspace{0.17em}⁣}{[{\displaystyle \frac{4x^3}{3}}+2x^2+x]}_0^2=\pi \text{\hspace{0.17em}⁣}({\displaystyle \frac{32}{3}}+8+2)=\pi \cdot {\displaystyle \frac{32+30}{3}}={\displaystyle \frac{62\pi }{3}}$.

A square homogeneous system has non-trivial (non-zero) solutions iff $\det (A)=0$.