Question

Докажите тождество, даю 100 баллов

Answer 1

Ответ:

$1)\ \ \boxed{\ 1+cos2a=2cos^2x\ \ ,\ \ 1-cos2x=2sin^2x\ }\\\\1-2cosa+cos2a=(1+cos2a)-2cosa=2cos^2a-2cosa=\\\\=2cosa\cdot (cosa-1)=-2cosa\cdot (1-cosa)=-2cosa\cdot 2sin^2\dfrac{a}{2}=-4cosa\cdot sin^2\dfrac{a}{2}\\\\-4cosa\cdot sin^2\dfrac{a}{2}=-4cosa\cdot sin^2\dfrac{a}{2}$

$2)\ \ \boxed{\ sin2a=2\, sina\cdot cosa\ \ ,\ \ cos(\dfrac{\pi}{2}\pm a)=sina\ \ ,\ \ cos(\dfrac{3\pi }{2}\pm a)=\pm sina}\\\\\\\dfrac{1-cos(4a-2\pi )+cos(4a-\dfrac{\pi}{2})}{1+cos(4a+\pi )+cos(4a+\dfrac{3\pi}{2})}=\dfrac{1+cos4a+sin4a}{1-cos4a+sin4a}=\\\\=\dfrac{2cos^22a+2\, sin2a\cdot cos2a}{2sin^22a+2\, sin2a\cdot cos2a}=\dfrac{2\, cos2a\cdot (cos2a+sin2a)}{2\, sin2a\cdot (sin2a+cos2a)}=\dfrac{cos2a}{sin2a}=ctg2a\\\\\\ctg2a=ctg2a$

$4)\ \ (sina+sin\beta )^2+(cosa+cos\beta )^2=\\\\=sin^2a+sin^2\beta +2sina\cdot sin\beta +cos^2a +cos^2\beta +2cosa\cdot cos\beta =\\\\=(\underbrace{sin^2a+cos^2a}_{1})+(\underbrace{sin^2\beta +cos^2\beta }_{1})+2(\underbrace{sina\cdot sina+sin\beta \cdot cos\beta }_{cos(a-\beta)})=\\\\=1+1+2\cdot cos(a-\beta )=2+2\cdot cos(a-\beta )=2\cdot (1+cos(a-\beta )\, )=\\\\=2\cdot 2\, cos^2\dfrac{a-\beta }{2}=4\, cos^2\dfrac{a-\beta }{2}\\\\\\4\, cos^2\dfrac{a-\beta }{2}=4\, cos^2\dfrac{a-\beta }{2}$

$\displaystyle \boxed{\ sinx\cdot cosy-cosx\cdot siny=sin(x-y)\ \ ,\ \ sinx+siny=2\cdot sin\dfrac{x+y}{2}\cdot cos\dfrac{x-y}{2}}$

$5)\ \ \displaystyle \Big(\frac{sin4a}{sina}-\frac{cos4a}{cosa}\Big)\Big(\frac{1}{sin3a}+\frac{1}{sina}\Big)=\\\\\\=\frac{sin4a\cdot cosa-cos4a\cdot sina}{sina\cdot cosa}\cdot \dfrac{sina+sin3a}{sina\cdot sin3a}=\frac{sin(4a-a)}{sina\cdot cosa}\cdot \frac{2\, sin2a\cdot cosa}{sina\cdot sin3a}=\\\\\\=\frac{sin3a}{sina\cdot cosa}\cdot \frac{2\cdot 2\, sina\cdot cosa\cdot cosa}{sina\cdot sin3a}=\frac{4\, cosa}{sina}=4\, ctga\\\\\\4\, ctga=4\, ctga$