Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
g,\\
\cos x.\cos y + \sin x.\sin y = \cos \left( {x - y} \right)\\
\Rightarrow G = \cos \frac{{5\pi }}{9}.\cos \frac{{7\pi }}{{18}} + \sin \frac{{5\pi }}{9}.\sin \frac{{7\pi }}{{18}} = \cos \left( {\frac{{5\pi }}{9} - \frac{{7\pi }}{{18}}} \right) = \cos \frac{\pi }{6} = \frac{{\sqrt 3 }}{2}\\
h,\\
\sin x.\cos y + \cos x.\sin y = \sin \left( {x + y} \right)\\
H = \sin \frac{{13\pi }}{4}.\cos \frac{{4\pi }}{7} + \sin \frac{{4\pi }}{7}.\cos \frac{{13\pi }}{4} = \sin \left( {\frac{{13\pi }}{4} + \frac{{4\pi }}{7}} \right) = \sin \frac{{107\pi }}{{28}} = - 0,532\\
2,\\
\sin \left( {\alpha + \frac{\pi }{6}} \right) - \cos \left( {\alpha - \frac{{2\pi }}{3}} \right)\\
= \left( {\sin \alpha .\cos \frac{\pi }{6} + \cos \alpha .\sin \frac{\pi }{6}} \right) - \left( {\cos \alpha .\cos \frac{{2\pi }}{3} + \sin \alpha .sin\frac{{2\pi }}{3}} \right)\\
= \left( {\frac{{\sqrt 3 }}{2}\sin \alpha + \frac{1}{2}\cos \alpha } \right) - \left( { - \frac{1}{2}\cos \alpha + \frac{{\sqrt 3 }}{2}\sin \alpha } \right)\\
= \cos \alpha = \frac{1}{3}\\
3,\\
\frac{\pi }{2} < \alpha < \pi \Rightarrow \cos \alpha < 0 \Rightarrow \cos \alpha = - \sqrt {1 - {{\sin }^2}\alpha } = - \frac{3}{5}\\
\pi < \beta < \frac{{3\pi }}{2} \Rightarrow \cos \beta < 0 \Rightarrow \cos \beta = - \sqrt {1 - {{\sin }^2}\beta } = - \frac{4}{5}\\
\cos \left( {\alpha + \beta } \right) = \cos \alpha .\cos \beta - sin\alpha .sin\beta = \frac{{24}}{{25}}\\
\cos \left( {\alpha - \beta } \right) = \cos \alpha .\cos \beta + \sin \alpha .\sin \beta = 0\\
\sin \left( {\alpha + \beta } \right) = \sin \alpha .cos\beta + cos\alpha .sin\beta = - \frac{7}{{25}}\\
\sin \left( {\alpha - \beta } \right) = \sin \alpha .cos\beta - \cos \alpha .\sin \beta = - 1
\end{array}\)
