Question

Помогите сделать срочно по математике

Answer 1

$1.;sin^2fracpi4-2cosfracpi3+3tg0^o=left(sinfracpi4right)^2-2cdotfrac12+3cdot0=left(frac1{sqrt2}right)^2-1=frac12-1=-frac12\\2.;sinalpha=frac5{13}\cosalpha=sqrt{1-sin^2alpha}=sqrt{1-frac{25}{169}}=sqrt{frac{144}{169}}=pmfrac{12}{13}\fracpi2<alpha<piRightarrowcosalpha<0\cosalpha=-frac{12}{13}\tgalpha=frac{sinalpha}{cosalpha}=frac5{13}:frac{12}{13}=frac5{13}cdotfrac{13}{12}=frac5{12}\ctgalpha=frac1{tgalpha}=frac{12}5=2frac25=2,4$

$3.;cos^2x-cos^4x+sin^4x=cos^2x(1-cos^2x)+sin^4x=cos^2xsin^2x+sin^4x=\=sin^2x(cos^2x+sin^2x)=sin^2x\\4.;sin^4alpha-cos^4alpha+cos^2alpha=sin^4alpha+cos^2alpha(1-cos^2alpha)=sin^4alpha+cos^2alphasin^2alpha=\=sin^2alpha(sin^2alpha+cos^2alpha)=sin^2alpha$