Question

Упростите 1-2sin2a +cos(4a-2п)​

Answer 1

$1-2sin2a+cos(4a-2\pi )=1-2sin2a+cos(2\pi -4a)=\\\\=1-2sin2a+cos(-4a)=1-2sin2a+cos4a=\\\\=(\underbrace {sin^2a+cos^2a}_{1})-2\cdot \underbrace {2\cdot sina\cdot cosa}_{sin2a}+(\underbrace {cos^22a-sin^22a}_{cos4a})=\\\\=(sin^2a-2\cdot sina\cdot cosa+cos^2a)-2\, sina\cdot cosa+(1-sin^22a-sin^22a)=\\\\=(sina-cosa)^2-sin2a+1-2sin^22a=\\\\=(sina-cosa)^2-(2sin^22a+sin2a-1)=\\\\=\Big[\ 2t^2+t-1=0\ ,\ t_1=-1\ ,\ t_2=1/2\ ,\ 2t^2+t-1=2(t+1)(t-\frac{1}{2})=\\\\=(t+1)(2t-1)\ \Big]=(sina-cosa)^2-(sin2a+1)(2sin2a-1)$

$ili:\ \ \ 1-2sin2a+cos(4a-2\pi )=1-2sin2a+cos(2\pi -4a)=\\\\=1-2sin2a+cos(-4a)=1-2sin2a+cos4a=\\\\=1-2sin2a+(\underbrace {cos^22a-sin^22a}_{cos4a})=1-2sin2a+(1-sin^22a-sin^22a)=\\\\=2-2sin2a-2sin^22a=-2(sin^22a+sin2a-1)=\\\\=\Big[\ t^2+t-1=0\ ,\ D=1+4=5\ ,\ t_{1,2}=\dfrac{-1\pm \sqrt5}{2}\ \Big]=\\\\=-2\, \Big(sin2a-\dfrac{-1-\sqrt5}{2}\Big)\Big(sin2a-\dfrac{-1+\sqrt5}{2}\Big)=\\\\=-2\, \Big(sin2a+\dfrac{1+\sqrt5}{2}\Big)\Big(sin2a+\dfrac{1-\sqrt5}{2}\Big)$