Question

тригонометрия срочно

Answer 1

$1); ; sinfrac{5pi }{4}=cosfrac{5pi }{4}=-frac{sqrt2}{2}; ; ,; ; tgfrac{5pi }{4}=ctgfrac{5pi }{4}=1\\sin(-900^circ)=0; ,; ; cos(-900^circ )=-1; ,; ; tg(-900^circ )=0; ,\ctg(-900^circ ); -; ne; syshestvyet\\2); ; cosa=-frac{sqrt2}{2}; ; ,; ; ain (pi ,frac{3pi}{2}); to ; sina<0; ,\\sina=-sqrt{1-cos^2a}=-sqrt{1-frac{2}{4}}=-frac{sqrt2}{2}\\cosbeta=frac{1}{2}; ; ,; ; beta in (0,frac{pi}{2}); to ; sinbeta >0; ,\\sinbeta =sqrt{1-cos^2beta }=sqrt{1-frac{1}{4}}=frac{sqrt3}{2}$

$cos(a+beta )=cosacdot cosbeta -sinacdot sinbeta =-frac{sqrt2}{2}cdot frac{1}{2}+frac{sqrt2}{2}cdot frac{sqrt3}{2}=frac{sqrt6-sqrt2}{4}\\3); a); 1-2sin^2frac{pi}{8}=cos(2cdot frac{pi}{8})=cosfrac{pi}{4}=frac{sqrt2}{2}\\b); ; sin105^circ =sin(60^circ +45^circ )=sin60^circ cdot cos45^circ +cos60^circ cdot sin45^circ =\\=frac{sqrt3}{2}cdot frac{sqrt2}{2}+frac{1}{2}cdot frac{sqrt2}{2}=frac{sqrt6+sqrt2}{4}\\c); ; ctgfrac{7pi}{4}=ctg(2pi -frac{pi }{4}=ctg(-frac{pi}{4})=-ctgfrac{pi}{4}=-frac{sqrt2}{2}$

$4); ; cos^4a+sin^4a-2cos^2a, sin^2a=(cos^2a-sin^2a)^2=(cos2a)^2=cos^22a; ;\\(1-cos^2a)cdot ctga=sin^2acdot frac{cosa}{sina}=sinacdot cosa$