Question

вообще некому помочь:((((((​

Answer 1

Ответ:

$6)\ \ cosa=-\dfrac{5}{13}\\\\sin^2a+cos^2a=1\ \ \ \Rightarrow\ \ \ sin^2a=1-cos^2a=1-\dfrac{25}{169}=\dfrac{144}{169}\ ,\\\\sina=\pm \sqrt{\dfrac{144}{169}}=\pm \dfrac{12}{13}\\\\a\in \Big(\pi \, ;\, \dfrac{3\pi }{2}\Big)\ \ \Rightarrow \ \ sina<0\ \ \ \Rightarrow \ \ sina=-\dfrac{12}{13}\\\\\\sin2a=2\cdot sina\cdot cosa=2\cdot \Big(-\dfrac{12}{13}\Big)\cdot \Big(-\dfrac{5}{13}\Big)=\dfrac{120}{169}\\\\cos2a=cos^2a-sin^2a=cos^2a-(1-cos^2a)=2cos^2a-1=2\cdot \dfrac{25}{169}-1=$

$=\dfrac{50}{169}-1=-\dfrac{119}{169}\\\\tg2a=\dfrac{sin2a}{cos2a}=\dfrac{120}{119}\\\\ctg2a=\dfrac{1}{tg2a}=\dfrac{119}{120}$

$1)\ \ \dfrac{sin2a}{cosa}-sina=\dfrac{2\cdot sina\cdot cosa}{cosa}-sina=2\cdot sina-sina=sina\\\\\\2)\ \ cos^2a-cos2a=cos^2a-(cos^2a-sin^2a)=cos^2a-cos^2a+sin^2a=sin^2a$