Show that $\sin(x + y)\sin(x - y) \equiv \sin^2 x - \sin^2 y$, for any \(x, y \in \mathbb{R}\

Question

Show that $\sin(x + y)\sin(x - y) \equiv \sin^2 x - \sin^2 y$, for any $x, y \in \mathbb{R}$. In triangle $ABC$, $a$, $b$ and $c$ are the opposite sides of the interior angles $A$, $B$ and $C$, respectively. Hence, show that $\frac{a^2 - b^2}{c^2} \equiv \frac{\sin(A - B)}{\sin C}$. Given further that $A$, $B$ and $C$ are acute angles and $b^2 - a^2 = ac$, use the result in (i), or otherwise, find the range of values of $A$ in terms of $\pi$.

Accepted Answer

(iii) $\frac{\pi }{6} < A < \frac{\pi }{4}$

Trigonometric Equations and Identities

Trigonometric Equations and Identities