Daily math problems for July 14, 2026 — Trigonometry, Logic, Linear Function & more

$\bullet$ 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:

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

The contrapositive of "$n^2$ even $\Rightarrow$ $n$ even" is "$n$ odd $\Rightarrow$ $n^2$ odd". A statement and its contrapositive always have the same truth value.

$m={\displaystyle \frac{11-2}{4-1}}={\displaystyle \frac{9}{3}}=3$.

Set $\sqrt{x^2+y^2}=\sqrt{(x-4{)}^2+y^2}$. Squaring: $x^2=(x-4{)}^2=x^2-8x+16$, so $8x=16$, $x=2$.

$P=\left(\genfrac{}{}{0px}{}{4}{2}\right)\text{\hspace{0.17em}⁣}{\left({\displaystyle \frac{1}{3}}\right)}^2\text{\hspace{0.17em}⁣}{\left({\displaystyle \frac{2}{3}}\right)}^2=6\cdot {\displaystyle \frac{1}{9}}\cdot {\displaystyle \frac{4}{9}}={\displaystyle \frac{24}{81}}={\displaystyle \frac{8}{27}}$.

For any $x\in [-1,1]$, $\arcsin x+\arccos x={\displaystyle \frac{\pi }{2}}$ (a standard identity, since $\arccos x={\displaystyle \frac{\pi }{2}}-\arcsin x$).

Middle coefficient: $\left(\genfrac{}{}{0px}{}{8}{4}\right)=70$.

${\sin }^2\theta =1-{\displaystyle \frac{4}{9}}={\displaystyle \frac{5}{9}}$, so $\sin \theta ={\displaystyle \frac{\sqrt{5}}{3}}$. Hence $\tan \theta ={\displaystyle \frac{\sin \theta }{\cos \theta }}={\displaystyle \frac{\sqrt{5}\mathrm{/}3}{2\mathrm{/}3}}={\displaystyle \frac{\sqrt{5}}{2}}$.

$e^{a+b}=e^a{e}^b$ ✓ (homomorphism). $e^x$ is strictly increasing (injective) and surjective onto $(0,\mathrm{\infty })$. Hence an isomorphism. (Inverse: $\ln$.)

Cauchy's equation with continuity gives $f(x)=cx$ for some constant $c$. Since $f(1)=3$, $c=3$, so $f(5)=15$.