Question

Помогите пожалуйста что то из этого решите. Особенно 5 или 6​

Answer 1

=$\displaystyle\bf\\5)\\\\\frac{Sin9x}{2Sin3x}+\frac{Cos9x}{2Cos3x} =\frac{Sin9x\cdot 2Cos3x+Cos9x\Cdot 2Sin3x}{2Sin3x\cdot 2Cos3x} =\\\\\\ =\frac{2\cdot(Sin9x\cdot Cos3x+Cos9x\Cdot Sin3x)}{2Sin6x} =\frac{Sin(3x+9x)}{Sin6x} =\\\\\\=\frac{Sin12x}{Sin6x}=\frac{2Sin6x Cos6x}{Sin6x}=2Cos6x\\\\\\x=\frac{\pi }{24} \\\\\\2Cos6x=2Cos\Big(6\cdot\frac{\pi }{24} \Big)=2Cos\frac{\pi }{4}=2\cdot\frac{\sqrt{2} }{2} =\sqrt{2}$

$\displaystyle\bf\\6)\\\\\frac{Sin^{2}\Big(\pi+ \alpha\Big)-Sin^{2} \Big( \dfrac{\pi }{2} +\alpha \Big)-Sin\Big(\pi- \alpha\Big) Sin\Big(2\pi+ \alpha\Big) }{tg\Big(\dfrac{\pi }{2}+\alpha \Big) tg\Big(\dfrac{3\pi }{2} -\alpha \Big) } =\\\\\\=\frac{Sin^{2} \alpha -Cos^{2}\alpha -Sin\alpha \cdot Sin\alpha }{-Ctg\alpha \cdot Ctg\alpha } = \frac{Sin^{2} \alpha -Cos^{2}\alpha -Sin^{2} \alpha }{-Ctg^{2} \alpha } =$

$\displaystyle\bf\\=\frac{Cos^{2}\alpha }{Ctg^{2}\alpha } =\frac{Cos^{2}\alpha }{\frac{Cos^{2} \alpha }{Sin^{2} \alpha } }=\frac{Cos^{2}\alpha \cdot Sin^{2}\alpha }{Cos^{2}\alpha } =Sin^{2} \alpha$