Страница 155 номер 519, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(2\cos^{2}\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) = 1 + \sin\alpha\)
\(2\cos^{2}\left(\dfrac{\pi}{4} - \dfrac{\alpha}{2}\right) = 2 \cdot \dfrac{1 + \cos(\frac{\pi}{2} - \alpha)}{2} =\)
\(= 1 + \cos\dfrac{\pi}{2}\cos\alpha + \sin\dfrac{\pi}{2}\sin\alpha = 1 + \sin\alpha;\)

2) \(2\sin^{2}\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) = 1 - \sin\alpha\)
\(2\sin^{2}\left(\dfrac{\pi}{4} - \dfrac{\alpha}{2}\right) = 1 - \cos\left(\dfrac{\pi}{2} - \alpha\right) =\)
\(= 1 - \cos\dfrac{\pi}{2}\cos\alpha - \sin\dfrac{\pi}{2}\sin\alpha =\)
\(= 1 - \sin\alpha;\)

3) \(\frac{3 - 4\cos 2\alpha + \cos 4\alpha}{3 + 4\cos 2\alpha + \cos 4\alpha} = \operatorname{tg}^{4}\alpha\)
\(\dfrac{3 - 4\cos 2\alpha + \cos 4\alpha}{3 + 4\cos 2\alpha + \cos 4\alpha} = \dfrac{2 + \cos^{2} 2\alpha + \sin^{2} 2\alpha - 4\cos 2\alpha + \cos^{2} 2\alpha - \sin^{2} 2\alpha}{2 + \cos^{2} 2\alpha + \sin^{2} 2\alpha + 4\cos 2\alpha + \cos^{2} 2\alpha - \sin^{2} 2\alpha} =\)
\(= \dfrac{2 - 4\cos 2\alpha + 2\cos^{2} 2\alpha}{2 + 4\cos 2\alpha + 2\cos^{2} 2\alpha} = \dfrac{2(1 - \cos 2\alpha)^{2}}{2(1 + \cos 2\alpha)^{2}} = \operatorname{tg}^{4}\alpha;\)

4) \(\frac{1 + \sin 2\alpha + \cos 2\alpha}{1 + \sin 2\alpha - \cos 2\alpha} = \operatorname{ctg}\alpha\)
\(\dfrac{1 + \sin 2\alpha + \cos 2\alpha}{1 + \sin 2\alpha - \cos 2\alpha} = \dfrac{\cos^{2}\alpha + \sin^{2}\alpha + 2\sin\alpha\cos\alpha + \cos^{2}\alpha - \sin^{2}\alpha}{\cos^{2}\alpha + \sin^{2}\alpha + 2\sin\alpha\cos\alpha - \cos^{2}\alpha + \sin^{2}\alpha} = \dfrac{2\cos^{2}\alpha + 2\sin\alpha\cos\alpha}{2\sin^{2}\alpha + 2\sin\alpha\cos\alpha} =\)
\(= \dfrac{2\cos\alpha(\cos\alpha + \sin\alpha)}{2\sin\alpha(\sin\alpha + \cos\alpha)} = \operatorname{ctg}\alpha.\)