Daily Math · 2026-06-29

Question 1

Find $\sin 75° - \sin 15°$.

Accepted Answer

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

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

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

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

$\bullet$ Evaluate:

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

Answer

$\dfrac{\sqrt{6}}{2}$

Answer

$\dfrac{\sqrt{2}}{2}$

Answer

$\dfrac{\sqrt{3}}{2}$

Answer

$\dfrac{1}{2}$

Answer

$1$

Answer

$\dfrac{\sqrt{2}}{4}$

Question 2

For $A = \begin{pmatrix} 1 & 0 \ 0 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 4 \ 6 & 8 \end{pmatrix}$, solve $XA = B$. Then $X$ equals:

Accepted Answer

$A^{-1} = \begin{pmatrix} 1 & 0 \ 0 & 1/2 \end{pmatrix}$. So $X = B \cdot A^{-1} = \begin{pmatrix} 2 \cdot 1 & 4 \cdot 1/2 \ 6 \cdot 1 & 8 \cdot 1/2 \end{pmatrix} = \begin{pmatrix} 2 & 2 \ 6 & 4 \end{pmatrix}$.

Answer

$\begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix}$

Answer

$\begin{pmatrix} 2 & 4 \ 3 & 4 \end{pmatrix}$

Answer

$\begin{pmatrix} 2 & 2 \ 6 & 4 \end{pmatrix}$

Answer

$\begin{pmatrix} 2 & 1 \ 6 & 2 \end{pmatrix}$

Question 3

For the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$, the eccentricity equals:

Accepted Answer

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

Answer

$\dfrac{3}{5}$

Answer

$\dfrac{4}{5}$

Answer

$\dfrac{5}{4}$

Answer

$\dfrac{4}{3}$

Problems & worked solutions

Problem 1 — Trigonometry

Problem 2 — Matrix Equations

Problem 3 — Conic Sections