Daily math problems for June 20, 2026 — Calculus, Quadratic Function, Trigonometry

$f(0)=-1<0$ and $f(1)=1>0$. Since $f$ is continuous, by the IVT there exists $c\in (0,1)$ with $f(c)=0$.

$f(2)=12-14+2=0$. So $f(2025)\cdot f(2)=f(2025)\cdot 0=0$, regardless of $f(2025)$.

$\bullet$ Multiply numerator and denominator by $2\sin 20\mathrm{°}$ and use $2\sin \theta \cos \theta =\sin 2\theta$:

$$\cos 20\mathrm{°}\cos 40\mathrm{°}\cos 80\mathrm{°}={\displaystyle \frac{\sin 40\mathrm{°}\cos 40\mathrm{°}\cos 80\mathrm{°}}{2\sin 20\mathrm{°}}}$$

$\bullet$ Apply the identity again:

$$={\displaystyle \frac{\sin 80\mathrm{°}\cos 80\mathrm{°}}{4\sin 20\mathrm{°}}}$$

$\bullet$ And once more:

$$={\displaystyle \frac{\sin 160\mathrm{°}}{8\sin 20\mathrm{°}}}$$

$\bullet$ Use $\sin 160\mathrm{°}=\sin (180\mathrm{°}-20\mathrm{°})=\sin 20\mathrm{°}$:

$$={\displaystyle \frac{\sin 20\mathrm{°}}{8\sin 20\mathrm{°}}}={\displaystyle \frac{1}{8}}$$