Question

2 и 4 номера (на изображение)

Answer 1

$2); ; ctga=sqrt2; ,; ; ain (pi ;2pi )\\1+ctg^2a=frac{1}{sin^2a}; ,; ; 1+(sqrt2)^2=frac{1}{sin^2a}; ,; ; 1+2=frac{1}{sin^2a}; ,\\sin^2a=frac{1}{3}; ,; ; 1-cos^2a=frac{1}{3}; ,; ; cos^2a=1-frac{1}{3}; ,; ; cos^2a=frac{2}{3}; ,\\cosa=pm sqrt{frac{2}{3}}\\ain (pi ;2pi ); ; Rightarrow \\a); ; cosa textless 0; ,; pri; ; ain (pi ;frac{3pi}{2}); ,; cosa=-sqrt{frac{2}{3}}=-frac{sqrt6}{3}\\b); ; cosa textgreater 0; ; pri; ; xin (frac{3pi}{2};2pi ); ; to ; ; cosa=+sqrt{frac{2}{3}}=frac{sqrt6}{3}$

$ctga=sqrt2 textgreater 0; ; to ; ; ain (frac{3pi}{2},2pi ); ; Rightarrow ; ; cosa textless 0; ,; cosa=-frac{sqrt6}{3}$

$frac{sqrt6}{cosa}=-frac{sqrt6}{sqrt6/3}=-3\\4); ; (sin^4frac{35pi }{24}-sin^4frac{25pi }{24})cdot sinfrac{25pi }{12}cdot sinfrac{25pi }{6}=\\=[; sin(frac{35pi}{24})=sin(pi +frac{11pi }{24})=-sinfrac{11pi }{24}; ,\\sinfrac{25pi }{24}=sin(pi +frac{pi }{24})=-sinfrac{pi }{24}; ]=\\=(sin^4frac{11pi }{24}- sin^4frac{pi }{24})cdot sin(2pi +frac{pi }{12})cdot sin(4pi +frac{pi }{6})=\\=(sinfrac{11pi }{24}-sinfrac{pi }{24})(sinfrac{11pi }{24}+sinfrac{11pi }{24}+sinfrac{pi }{24})(sin^2frac{11pi}{24}+sin^2frac{pi}{24})cdot \\cdot sinfrac{pi}{12}cdot sinfrac{pi}{6}=$

$=2, sinfrac{5pi }{24}cdot cosfrac{pi }{4}cdot 2, sinfrac{pi }{4}cdot cosfrac{5pi }{24}cdot frac{sqrt3-1}{2sqrt2}cdot frac{1}{2}=\\=2cdot sinfrac{5pi }{24}cdot cosfrac{5pi}{24}cdot frac{sqrt2}{2}cdot frac{sqrt2}{2}cdot frac{sqrt3-1}{2sqrt2}=\\=sinfrac{5pi }{12}cdot frac{sqrt3-1}{4sqrt2}=sin(frac{pi}{2}-frac{pi}{12})cdot frac{sqrt2(sqrt3-1)}{4sqrt2cdot sqrt2}=\\=cosfrac{pi }{12}cdot frac{sqrt6-sqrt2}{8}=frac{sqrt3+1}{2sqrt2}cdot frac{sqrt6-sqrt2}{8}=frac{sqrt6+sqrt2}{4}cdot frac{sqrt6-sqrt2}{8}=$

$frac{(sqrt6)^2-(sqrt2)^2}{4cdot 8}=frac{6-2}{4cdot 8}=frac{4}{4cdot 8}=frac{1}{8}$