Đáp án + Giải thích các bước giải:

$\textbf{a}\bigg)$

$\textbf{VT} = \sin (60^\circ + \alpha) - \sin (60^\circ - \alpha)$ 

$= \sin 60^\circ \cos \alpha + \sin \alpha \cos 60^\circ - (\sin 60^\circ \cos \alpha - \sin \alpha \cos 60^\circ)$

$= \sin 60^\circ \cos \alpha + \sin \alpha \cos 60^\circ - \sin 60^\circ \cos \alpha + \sin \alpha \cos 60^\circ$

$= 2\cos 60^\circ \sin \alpha$

$= 2 \cdot \dfrac{1}{2} \sin \alpha$

$= \sin\alpha = \textbf{VP}$

$\textbf{b}\bigg)$

$\textbf{VT} = \sin^4 \alpha + \cos^4 \alpha$

$= (\sin^2 \alpha + \cos^2 \alpha)^2 - 2\sin^2 \alpha \cos^2 \alpha$

$= 1^2 - \dfrac{1}{2}\sin^2 2\alpha$

$= 1 - \dfrac{1}{4}(2\sin^2 2\alpha)$

$= 1 - \dfrac{1}{4} (2\sin^2 2\alpha - 1 +1)$

$= 1 - \dfrac{1}{4} (2\sin^2 2\alpha - 1) - \dfrac{1}{4}$

$= \dfrac{3}{4} + \dfrac{1}{4} \cos 4\alpha = \textbf{VP}$

$\textbf{c}\bigg)$

$\textbf{VT} = \dfrac{\sin 3x \cos 2x + \sin x \cos 6x}{\sin 4x}$

$= \dfrac{\frac{1}{2}[\sin (3x + 2x) + \sin (3x - 2x)] + \frac{1}{2}[\sin (x + 6x) + \sin (x - 6x)]}{\sin 4x}$

$= \dfrac{\sin 5x + \sin x + \sin 7x + \sin (-5x)}{2\sin 4x}$

$= \dfrac{\sin 5x + \sin x + \sin 7x - \sin 5x}{2 \sin 4x}$

$= \dfrac{\sin x + \sin 7x}{2 \sin 4x}$

$= \dfrac{2 \sin \frac{x + 7x}{2} \cos \frac{x - 7x}{2}}{2 \sin 4x}$

$= \dfrac{\sin 4x \cos (-3x)}{\sin 4x}$

$= \cos (-3x)$

$= \cos 3x = \textbf{VP}$

$\textbf{d}\bigg)$

$\textbf{VT} = \dfrac{\cos (\alpha - \beta)}{\cos (\alpha + \beta)}$

$= \dfrac{\cos \alpha \cos \beta + \sin \alpha \sin\beta}{\cos \alpha \cos \beta - \sin \alpha \sin\beta}$

$= \dfrac{\cos \alpha \cos \beta \big(1 + \frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta}\big)}{\cos \alpha \cos \beta \big(1 - \frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta}\big)}$

$= \dfrac{1 + \frac{\sin \alpha}{\cos \alpha} \cdot \frac{\sin \beta}{\cos \beta}}{1 - \frac{\sin \alpha}{\cos \alpha} \cdot \frac{\sin \beta}{\cos \beta}}$

$= \dfrac{1 + \tan \alpha \tan \beta}{1 - \tan \alpha \tan \beta} = \textbf{VP}$