Question

Ля ребят помогите!!!дам много баллов

Answer 1

$1) \ \cos \alpha = \dfrac{3}{5};\ \cos \beta = -\dfrac{7}{25} ; \ \dfrac{3\pi}{2} < \alpha < 2\pi; \ \pi < \beta < \dfrac{3\pi}{2}$

$\sin(\alpha - \beta )= \sin\alpha \cos\beta - \sin\beta \cos\alpha$

$\sin^{2}\alpha + \cos^{2}\alpha = 1;\\\sin^{2}\alpha = 1 - \cos^{2}\alpha \\\sin \alpha = -\sqrt{1 - \cos^{2}\alpha} = -\sqrt{1 - \bigg( \dfrac{3}{5} \bigg)^{2}}= -\sqrt{1 - \dfrac{9}{25} }= -\sqrt{\dfrac{16}{25} } = -\dfrac{4}{5}$

$\sin \beta = -\sqrt{1 - \cos^{2}\beta} = -\sqrt{1 - \bigg( \dfrac{7}{25} \bigg)^{2}}= -\sqrt{1 - \dfrac{49}{625} }= -\sqrt{\dfrac{576}{625} } = -\dfrac{24}{25}$

$\sin(\alpha - \beta )= -\dfrac{4}{5} \cdot \bigg(-\dfrac{7}{25}\bigg) - \bigg(-\dfrac{24}{25}\bigg) \cdot \dfrac{3}{5} = \dfrac{28}{125} + \dfrac{72}{125} = \dfrac{100}{125} = 0,8$

Ответ: 0,8.

$2) \ \cos \bigg(\arcsin \dfrac{15}{17} \bigg) + \arccos \bigg(-\dfrac{12}{13}\bigg) = \sqrt{1 - \bigg( \dfrac{15}{17} \bigg)^{2}} + \pi - \arccos \dfrac{12}{13} =\\\\= \dfrac{8}{17} + \pi - \arccos \dfrac{12}{13}$

Ответ: $\dfrac{8}{17} + \pi - \arccos \dfrac{12}{13}$

$3) \ \dfrac{\sin \bigg( \dfrac{\pi}{4} + \alpha \bigg) - \cos \bigg( \dfrac{\pi}{4} + \alpha \bigg)}{\sin \bigg( \dfrac{\pi}{4} + \alpha \bigg) + \cos \bigg( \dfrac{\pi}{4} + \alpha \bigg)} =\\\\= \dfrac{\sin\dfrac{\pi}{4} \cos \alpha + \cos\dfrac{\pi}{4}\sin\alpha - \cos\dfrac{\pi}{4}\cos \alpha + \sin\dfrac{\pi}{4}\sin\alpha}{\sin\dfrac{\pi}{4} \cos \alpha + \cos\dfrac{\pi}{4}\sin\alpha + \cos\dfrac{\pi}{4}\cos \alpha - \sin\dfrac{\pi}{4}\sin\alpha} =$

$\dfrac{\dfrac{\sqrt{2}}{2} \cos \alpha + \dfrac{\sqrt{2}}{2}\sin\alpha -\dfrac{\sqrt{2}}{2}\cos \alpha + \dfrac{\sqrt{2}}{2}\sin\alpha}{\dfrac{\sqrt{2}}{2} \cos \alpha + \dfrac{\sqrt{2}}{2}\sin\alpha + \dfrac{\sqrt{2}}{2}\cos \alpha - \dfrac{\sqrt{2}}{2}\sin\alpha} = \dfrac{\sqrt{2}\sin\alpha }{\sqrt{2}\cos \alpha } = \dfrac{\sin\alpha }{\cos \alpha } = \text{tg} \ \alpha$

Тождество доказано

$4) \ \text{tg} \bigg(\dfrac{\pi}{3} + \alpha \bigg) - \text{tg} \bigg(\dfrac{\pi}{3} - \alpha \bigg) = \dfrac{\sin\bigg(\dfrac{\pi}{3} + \alpha \bigg)}{\cos\bigg(\dfrac{\pi}{3} + \alpha \bigg)} -\dfrac{\sin\bigg(\dfrac{\pi}{3} - \alpha \bigg)}{\cos\bigg(\dfrac{\pi}{3} - \alpha \bigg)} =$

$= \dfrac{\sin\bigg(\dfrac{\pi}{3} + \alpha \bigg)\cos\bigg(\dfrac{\pi}{3} - \alpha \bigg) - \sin\bigg(\dfrac{\pi}{3} - \alpha \bigg)\cos\bigg(\dfrac{\pi}{3} + \alpha \bigg)}{\cos\bigg(\dfrac{\pi}{3} + \alpha \bigg)\cos\bigg(\dfrac{\pi}{3} - \alpha \bigg)} =$

$= \dfrac{\sin \bigg(\dfrac{\pi}{3} + \alpha - \dfrac{\pi}{3} + \alpha \bigg)}{\dfrac{1}{2}\bigg(\cos \bigg(\dfrac{\pi}{3} + \alpha - \dfrac{\pi}{3} + \alpha \bigg) + \cos \bigg(\dfrac{\pi}{3} +\alpha + \dfrac{\pi}{3} - \alpha \bigg) \bigg) } =$

$= \dfrac{2\sin 2\alpha }{\cos 2\alpha +\cos\dfrac{2\pi}{3} } =\dfrac{2\sin2\alpha }{\cos2\alpha - \dfrac{1}{2} } = \dfrac{4\sin \alpha \cos \alpha }{2\cos^{2}\alpha - 1 - \dfrac{1}{2} } =\dfrac{4\dfrac{\sin \alpha }{\cos\alpha } \cdot \cos^{2}\alpha }{\dfrac{2}{1 + \text{tg}^{2}\alpha } - 1 \dfrac{1}{2} } =$

$= \dfrac{\dfrac{4\text{tg}\alpha }{1 + \text{tg}^{2}\alpha } }{\dfrac{1 - 3\text{tg}^{2}\alpha }{2 + 2 \text{tg}^{2}\alpha } } = \dfrac{4\text{tg}\alpha (2 + 2 \text{tg}^{2}\alpha)}{(1 + \text{tg}^{2}\alpha)(1 - 3\text{tg}^{2}\alpha)} =\dfrac{8\text{tg}\alpha (1 + \text{tg}^{2}\alpha)}{(1 + \text{tg}^{2}\alpha)(1 - 3\text{tg}^{2}\alpha)} =\\\\\\= \dfrac{8\text{tg}\alpha }{1 - 3\text{tg}^{2}\alpha}$

Тождество доказано.