Daily math problems for May 27, 2026 — Trigonometry, Polynomials, Calculus

$\bullet$ Apply the difference-to-product identity:

$$\sin A-\sin B=2\cos \frac{A+B}{2}\sin \frac{A-B}{2}$$

$\bullet$ With $A=75\mathrm{°}$, $B=15\mathrm{°}$ the half-sum is $45\mathrm{°}$ and half-difference is $30\mathrm{°}$:

$$\sin 75\mathrm{°}-\sin 15\mathrm{°}=2\cos 45\mathrm{°}\sin 30\mathrm{°}$$

$\bullet$ Evaluate:

$$2\cdot \frac{\sqrt{2}}{2}\cdot \frac{1}{2}=\frac{\sqrt{2}}{2}$$

$P(1)=1+2-1+5=7$.

The integral of a non-negative continuous function over $[a,b]$ is non-negative (representing the area under the curve).