Question

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Answer 1

$1^o\a);sin113^ocos67^o+sin67^ocos113^o=sin(113^o+67^o)=sin180^o=0\b);cos74^ocos29^o+sin74^ocos61^o=frac12left(cos45^o+cos103^o+sin13^o+sin135^oright)=;=frac12left(cos45^o+sin135^o+cos(fracpi2+13^o)+sin13^oright)=\=frac12left(frac{sqrt2}2+frac{sqrt2}2-sin13^o+sin13^oright)=frac{sqrt2}2$

$2^o\2cos^23alpha tg3alpha=5cos^23alphacdotfrac{sin3alpha}{cos3alpha}=5cos3alphasin3alpha=2,5cdotsin6alpha\\3.\sin^2alpha+cos^alpha=1Rightarrowcosalpha=sqrt{1-sin^2alpha}\cosalpha=sqrt{1-0,64}=sqrt{0,36}=pm0,6\fracpi2<alphapiRightarrowcosalpha<0\cosalpha=-0,6$

$4.\frac{sin(x+45^o)+sin(x-45^o)}{sin(x+45^o)-sin(x-45^o)}=frac{sin xcos45^o+cos xsin45^o+sin xcos45^o-cos xsin45^o}{sin xcos45^o+cos xsin45^o-sin xcos45^o+cos xsin45^o}=\=frac{2sin xcos45^o}{2cos xsin45^o}=frac{sqrt2sin x}{sqrt2cos x}=tgx$

$5.\cosalpha=frac23\sinalpha=sqrt{1-cos^2alpha}=sqrt{1-frac49}=sqrt{frac59}=pmfrac{sqrt5}3\270^o<alpha<360^oRightarrowsinalpha<0\sinalpha=-frac{sqrt5}3\sinbeta=frac13\cosbeta=sqrt{1-frac19}=sqrt{frac89}=pmfrac{2sqrt2}3\90^o<beta<180^oRightarrowcosbeta<0\cosbeta=$

$sin(2alpha+beta)=sin2alphacosbeta+cos2alphasinbeta=\=2sinalphacosalphacosbeta+(1-sin^2alpha)sinbeta=\=2cdotleft(-frac{sqrt5}3right)cdotfrac23cdotleft(-frac{2sqrt2}3right)+left(1-frac59right)cdotfrac13=frac{8sqrt{10}+4}{27}$