Question

Решите пожалуйста 1,2,3. Благодарю.

Answer 1

$1); ; cos alpha =frac{9}{11}; ; ,; ; frac{3pi }{2} leq alpha leq 2pi \\sin alpha =-sqrt{1-cos^2 alpha }=sqrt{1-frac{81}{121}}=-sqrt{frac{40}{121}}=-frac{2sqrt{10}}{11}\\tg alpha = frac{sin alpha }{cos alpha } =-frac{2sqrt{10}}{9}\\2); sin alpha =frac{2}{7}; ,; ; frac{pi }{2} leq alpha leq pi \\cos alpha =-sqrt{1-sin^2 alpha }=-sqrt{1-frac{4}{49}}=-sqrt{frac{45}{49}}=-frac{3sqrt{5}}{7}\\tg alpha =-frac{2}{3sqrt5}=-frac{2sqrt5}{15}$

$3); ; tg alpha =7; ,; ; 0 leq alpha leq frac{pi }{2}\\cos^2 alpha = frac{1}{1+tg^2 alpha } =frac{1}{1+49}= frac{1}{50} \\cos alpha =frac{1}{5sqrt2}=frac{sqrt2}{10}\\sin alpha =+sqrt{1-cos^2 alpha }=sqrt{1-frac{2}{100}}=sqrt{frac{98}{100}}=frac{7sqrt2}{10}\\oo$

Answer 2

1) Sin²a = 1 - Cos²a = 1 - 81/121 =40/121 ⇒
⇒Sin a = - 2√10/11( а∈IV четв)
      tg a = Sin a : Cos a = -2√10/11 : 9/11= -2√10/9
2) Сos²a = 1 - Sin²a = 1 - 4/49 = 45/49⇒
⇒Cos a = -3√5/7(a∈II четв.)
     tg a = Sin a : Cos a = 2/7:-3√5/7 = -2√5/15
3) 1 + tg² a = 1/Cos²a
     1 + 49 = 1/Cos²a
      50 = 1/Сos²а ⇒Cos²a = 1/50⇒Сos a = √2/10
Sin²a = 1 - Cos²a = 1 - 2/100 = 98/100⇒ Sin a = 7√2/10