Daily math problems for June 29, 2026 — Trigonometry, Matrices, Analytic Geometry

$\bullet$ 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:

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

$A^{-1}=\left(\begin{array}{cc}1& 0\\ 0& 1\mathrm{/}2\end{array}\right)$. So $X=B\cdot A^{-1}=\left(\begin{array}{cc}2\cdot 1& 4\cdot 1\mathrm{/}2\\ 6\cdot 1& 8\cdot 1\mathrm{/}2\end{array}\right)=\left(\begin{array}{cc}2& 2\\ 6& 4\end{array}\right)$.

$a=5$, $b=3$, $c=\sqrt{25-9}=4$. So $e={\displaystyle \frac{4}{5}}$.