Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
13,\\
\cos x.\sin y - \sin x.\cos y = \sin \left( {y - x} \right)\\
\cos \left( {x - y} \right) = \cos x.\cos y + \sin x.\sin y\\
\dfrac{{\cos x.\sin \left( {x - 3} \right) - \sin x.\cos \left( {x - 3} \right)}}{{\cos \left( {3 - \dfrac{\pi }{6}} \right) - \dfrac{1}{2}\sin 3}}\\
= \dfrac{{\sin \left[ {\left( {x - 3} \right) - x} \right]}}{{\cos 3.\cos \dfrac{\pi }{6} + \sin 3.\sin \dfrac{\pi }{6} - \dfrac{1}{2}\sin 3}}\\
= \dfrac{{\sin \left( { - 3} \right)}}{{\dfrac{{\sqrt 3 }}{2}\cos 3 + \dfrac{1}{2}\sin 3 - \dfrac{1}{2}\sin 3}}\\
= \dfrac{{ - \sin 3}}{{\dfrac{{\sqrt 3 }}{2}\cos 3}}\\
= \dfrac{{ - 2\tan 3}}{{\sqrt 3 }}\\
14,\\
\sin x - \sin y = 2\cos \dfrac{{x + y}}{2}.\sin \dfrac{{x - y}}{2}\\
\sin 20^\circ + 2\sin 40^\circ - \sin 100^\circ \\
= \left( {\sin 20^\circ - \sin 100^\circ } \right) + 2\sin 40^\circ \\
= 2.\cos \dfrac{{20^\circ + 100^\circ }}{2}.sin\dfrac{{20^\circ - 100^\circ }}{2} + 2\sin 40^\circ \\
= 2.\cos 60^\circ .\sin \left( { - 40^\circ } \right) + 2\sin 40^\circ \\
= - 2\sin 40^\circ .\cos 60^\circ + 2\sin 40^\circ \\
= 2\sin 40^\circ \left( { - \cos 60^\circ + 1} \right)\\
= 2\sin 40^\circ .\dfrac{1}{2}\\
= \sin 40^\circ
\end{array}\)
