Упр.10 ГДЗ Колмогоров 10-11 класс (Алгебра)
а) $$\sin \alpha=\frac45,\quad \cos \beta=-\frac5{13},\quad \frac{\pi}{2}<\alpha<\pi,\quad \frac{\pi}{2}<\beta<\pi.$$
Так как $$\frac{\pi}{2}<\alpha<\pi,$$ то $$\cos \alpha<0,$$ и
$$\cos \alpha=-\sqrt{1-\sin^2\alpha}=-\sqrt{1-\left(\frac45\right)^2}=-\sqrt{1-\frac{16}{25}}=-\frac35.$$
Так как $$\frac{\pi}{2}<\beta<\pi,$$ то $$\sin \beta>0,$$ и
$$\sin \beta=\sqrt{1-\cos^2\beta}=\sqrt{1-\left(-\frac5{13}\right)^2}=\sqrt{1-\frac{25}{169}}=\sqrt{\frac{144}{169}}=\frac{12}{13}.$$
Тогда
$$\sin 2\alpha=2\sin \alpha\cos \alpha=2\cdot \frac45\cdot \left(-\frac35\right)=-\frac{24}{25},$$
$$\cos 2\beta=\cos^2\beta-\sin^2\beta=\left(-\frac5{13}\right)^2-\left(\frac{12}{13}\right)^2=\frac{25}{169}-\frac{144}{169}=-\frac{119}{169},$$
$$\sin(\alpha-\beta)=\sin \alpha\cos \beta-\cos \alpha\sin \beta=\frac45\cdot \left(-\frac5{13}\right)-\left(-\frac35\right)\cdot \frac{12}{13}=\frac{16}{65},$$
$$\cos(\alpha+\beta)=\cos \alpha\cos \beta-\sin \alpha\sin \beta=\left(-\frac35\right)\cdot \left(-\frac5{13}\right)-\frac45\cdot \frac{12}{13}=-\frac{33}{65}.$$
б) $$\cos \alpha=0{,}6,\quad \sin \beta=-\frac8{17},\quad \frac{3\pi}{2}<\alpha<2\pi,\quad \pi<\beta<\frac{3\pi}{2}.$$
Так как $$\frac{3\pi}{2}<\alpha<2\pi,$$ то $$\sin \alpha<0,$$ и
$$\sin \alpha=-\sqrt{1-\cos^2\alpha}=-\sqrt{1-(0{,}6)^2}=-\sqrt{1-0{,}36}=-0{,}8.$$
Так как $$\pi<\beta<\frac{3\pi}{2},$$ то $$\cos \beta<0,$$ и
$$\cos \beta=-\sqrt{1-\sin^2\beta}=-\sqrt{1-\left(-\frac8{17}\right)^2}=-\sqrt{1-\frac{64}{289}}=-\sqrt{\frac{225}{289}}=-\frac{15}{17}.$$
Тогда
$$\sin 2\alpha=2\sin \alpha\cos \alpha=2\cdot (-0{,}8)\cdot 0{,}6=-0{,}96,$$
$$\cos 2\beta=\cos^2\beta-\sin^2\beta=\left(-\frac{15}{17}\right)^2-\left(-\frac8{17}\right)^2=\frac{225}{289}-\frac{64}{289}=\frac{161}{289},$$
$$\sin(\alpha-\beta)=\sin \alpha\cos \beta-\cos \alpha\sin \beta=(-0{,}8)\cdot \left(-\frac{15}{17}\right)-0{,}6\cdot \left(-\frac8{17}\right)=\frac{84}{85},$$
$$\cos(\alpha+\beta)=\cos \alpha\cos \beta-\sin \alpha\sin \beta=0{,}6\cdot \left(-\frac{15}{17}\right)-(-0{,}8)\cdot \left(-\frac8{17}\right)=-\frac{77}{85}.$$
Ответ
а) $$\sin 2\alpha=-\frac{24}{25},\ \cos 2\beta=-\frac{119}{169},\ \sin(\alpha-\beta)=\frac{16}{65},\ \cos(\alpha+\beta)=-\frac{33}{65}.$$
б) $$\sin 2\alpha=-0{,}96,\ \cos 2\beta=\frac{161}{289},\ \sin(\alpha-\beta)=\frac{84}{85},\ \cos(\alpha+\beta)=-\frac{77}{85}.$$
