Question

срочноооооооооо!!!!​

Answer 1

α - 3 четверть.

sinα= √1-25/169= √144/169

sinα= -12/13

tgα=sinα/cosα, tgα= 12/5

ctgα=5/12

1) sin2α=2sinα•cosα=2• (-12/13) • (-5/13)= 2 • 60/169= 120/169

2) cos2α=cos^2α-sin^2α= 25/169 - 144/169 = -119/169

3)tg2α= 2•tgα/ 1-tg^2α=2 • 12/5 / 1-144/25= 24/5 / (-119/25)= - 24/5 • 25/119 = -24 • 5/119= -120/119= -1 1/119

4) ctg2α= ctg^2α -1/ 2•ctgα= 25/144 - 1/ 2•5/12= -119/144 / 5/6= - 119/144 • 6/5 = -714/720

Answer 2

Ответ:

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

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