Question

Помогите пожалуйста, с решением

Answer 1

$\displaystyle\bf\\4)\\\\Cos^{2} 15^\circ-Sin^{2} 15^\circ=Cos(2\cdot 15^\circ)=Cos30^\circ=\frac{\sqrt{3} }{2} \\\\\\Otvet \ : \ b) \ \ \frac{\sqrt{3} }{2} \\\\5)\\\\Cos39^\circ Cos21^\circ-Sin39^\circ Sin21^\circ=Cos(39^\circ+21^\circ)=\\\\\\=Cos60^\circ=\frac{1}{2} \\\\\\Otvet \ : \ c) \ \ \frac{1}{2} \\\\6)\\\\2Sin\frac{\pi }{6} -3Cos\Big(-\frac{\pi }{3} \Big)=2Sin\frac{\pi }{6} -3Cos\frac{\pi }{3} =$

$\displaystyle\bf\\=2\cdot\frac{1}{2} -3\cdot\frac{1}{2}=1-\frac{3}{2} =-\frac{1}{2} \\\\\\Otvet \ : \ b) \ \ -\frac{1}{2}$