Question

Формули подвійного аргументу 1-4 ,помогите хоть началом

Answer 1

$1)\quad Formyla:\quad tg \frac{ \alpha }{2} =\frac{sin \alpha }{1+cos \alpha } = \frac{1-cos \alpha }{sin \alpha } \quad \star \\\\ \frac{sin2 \alpha }{1+cos2 \alpha } \cdot \frac{cos \alpha }{1+cos \alpha } \cdot \frac{cos\frac{ \alpha }{2}}{1+cos\frac{ \alpha }{2}} =tg \alpha \cdot \frac{cos \alpha }{1+cos \alpha } \cdot \frac{cos\frac{ \alpha }{2}}{1+cos\frac{ \alpha }{2}} =$

 $= \frac{sin \alpha }{cos \alpha }\cdot \frac{cos \alpha }{1+cos \alpha } \cdot \frac{cos\frac{ \alpha }{2}}{1+cos\frac{ \alpha }{2}} =$ 

$= \frac{sin \alpha \alpha }{1+cos \alpha } \cdot \frac{cos\frac{ \alpha }{2}}{1+cos\frac{ \alpha }{2}} =tg\frac{ \alpha }{2}\cdot \frac{cos\frac{ \alpha }{2}}{1+cos\frac{ \alpha }{2}} = \frac{sin\frac{ \alpha }{2}}{1+cos\frac{ \alpha }{2}} =tg\frac{ \alpha }{4}$

$2)\quad \frac{1-sin36^\circ }{cos36^\circ } = \frac{1-sin(90^\circ -54^\circ )}{cos(90^\circ -54^\circ )} =[\, \frac{1-cos \alpha }{sin \alpha }=tg\frac{ \alpha }{2}\, ]=\\\\=\frac{1-cos54^\circ }{sin54^\circ}=tg\frac{54^\circ }{2}=tg27^\circ \\\\3)\quad tg(\frac{\pi}{4}+\frac{ \alpha }{2})\cdot \frac{1-sin \alpha }{cos \alpha } =tg(\frac{\pi}{4}+\frac{ \alpha }{2})\cdot \frac{1-cos(\frac{\pi}{2}- \alpha )}{sin(\frac{\pi}{2}- \alpha )} =$

$=tg(\frac{\pi}{4}+\frac{ \alpha }{2})\cdot tg(\frac{\pi}{4}-\frac{ \alpha }{2})=tg(\frac{\pi}{4}+\frac{ \alpha }{2})\cdot ctg(\frac{\pi}{2}-(\frac{\pi}{4}-\frac{ \alpha }{2}))=\\\\=tg(\frac{\pi}{4}+\frac{ \alpha }{2})\cdot ctg(\frac{\pi}{4}+\frac{ \alpha }{2})=1$

$4)\quad \frac{tg(\frac{5\pi}{4}- \alpha )(1+sin2 \alpha )}{cos(\frac{5\pi}{2}-2 \alpha )} = \frac{tg(\pi +\frac{\pi}{4}- \alpha )(1+sin2 \alpha )}{cos(2\pi+\frac{\pi}{2}-2 \alpha )} =\\\\= \frac{tg(\frac{\pi}{4}- \alpha )(1+sin2 \alpha )}{sin2 \alpha } =\frac{tg(\frac{\pi}{4}- \alpha )(1+cos (\frac{\pi}{2}-2\alpha ))}{2sin 2\alpha }=$

$= \frac{tg(\frac{\pi}{4}- \alpha )\cdot 2cos^2(\frac{\pi}{4}- \alpha )}{sin2 \alpha } = \frac{\frac{sin(\frac{\pi}{4}- \alpha )}{cos(\frac{\pi}{4}- \alpha )}\cdot 2cos^2(\frac{\pi}{4}- \alpha )}{sin2 \alpha } =\\\\= \frac{2sin(\frac{\pi}{4}- \alpha )\cdot cos(\frac{\pi}{4}- \alpha )}{sin2 \alpha } = \frac{sin(\frac{\pi}{2}-2 \alpha )}{sin2 \alpha }= \frac{cos2 \alpha }{sin2 \alpha } =ctg2 \alpha$