Question

1) известно что:
2) упростите выражения 
3) докажите тождество

Answer 1

$sin(frac{pi}{2}+a)=-frac{1}{2}\ sin(30+a)\ \ sin(frac{pi}{2}+a)=cosa\ cosa=-frac{1}{2}\ a=-frac{2pi}{3}+2pi*n=frac{4pi}{3}\\ sin(30+240)=sin(270)=-1\\ 2cos^2a-(tga*cosa)^2-(ctga*sina)^2=\ 2cos^2a-(frac{sina}{cosa}*cosa)^2-(frac{cosa}{sina}*sina)^2=\ 2cos^2a-sin^2a-cos^2a=\ cos^2a-sin^2a=cos2a$

$frac{sina+sin3a}{cosa+cos3a}*(1+cos4a)=\ frac{2sin2a*cosa}{2cos2a*cosa}*(1+cos4a)=\ frac{sin2a}{cos2a}*(1+cos(2*2a))=\ frac{sin2a}{cos2a}*(2cos^2(2a))=\ 2sin2a*cos2a=sin4a\ \ ctg2a*frac{2tga}{1+tg^2a}=\ ctg2a*sin2a=\ frac{cos2a}{sin2a}*sin2a=cos2a$

Answer 2

1)
$sin(frac{pi}{2}+alpha)=-frac{1}{2}; \ pi<alpha<frac{3pi}{2}; alphain(pi;frac{2pi}{3});\ sin(30^0+alpha)-?;\ frac{pi}{2}+alpha=(-1)^narcsin(-frac{1}{2})+pi n; nin Z\ alpha=-(-1)^narcsin(frac{1}{2})-frac{pi}{2}+pi n;\ alpha=(-1)^{n+1}cdotfrac{pi}{6}-frac{pi}{2}+pi n;\ n=2: alpha=-frac{pi}{6}-frac{pi}{2}+2pi=frac{-pi-3pi+12pi}{6}=frac{8pi}{6}=frac{4pi}{3}in(pi;frac{2pi}{3});$
$sin(30^0+alpha)=sin(frac{pi}{6}+frac{4pi}{3})=sin(frac{pi+8pi}{6})=\ =sinfrac{9pi}{6}=sin{frac{3pi}{2}}-1;\ or: sinfrac{pi}{6}cosfrac{4pi}{3}+sin{frac{4pi}{3}}cosfrac{pi}{6}=\ =frac{1}{2}cdot(-frac{1}{2})+(-frac{sqrt{3}}{2})cdotfrac{sqrt{3}}{2}=-frac{1}{4}-frac{3}{4}=-frac{4}{4}=-1$

$sin(30^0+alpha)=-1;$

2)
a)
$2cos^2alpha-(tgalphacdotcosalpha)^2-(ctgalphacdotsinalpha)^2=\ =2cos^2alpha-left(left(frac{sinalpha}{cosalpha}cdotcosalpharight)^2+left(frac{cosalpha}{sinalpha}cdot sinalpharight)^2right)=\ =2cos^2alpha-left(sin^2alpha+cos^2alpharight)=2cos^2alpha-1=cos2alpha;$
б)
$frac{sinalpha+sin3alpha}{cosalpha+cos3alpha}cdot(1+cos4alpha)=\ =frac{sinalpha+sin(alpha+2alpha)}{cosalpha+cos(alpha+2alpha)}cdot(1+2cos^22alpha-1)=\ =frac{sinalpha+sinalphacos2alpha+cosalphasin2alpha}{cosalpha+cosalphacos2alpha-sinalphasin2alpha}cdot2cos^22alpha=\ ==frac{sinalpha+sinalphacos2alpha+2cos^2alphasinalpha}{cosalpha+cosalphacos2alpha-2sin^2alphacosalpha}cdot2cos^22alpha=$
$=frac{sinalpha(1+cos2alpha+2cos^2alpha)}{cosalpha(1+cos2alpha-2sin^2alpha)}cdot2cos^22alpha=\ frac{sinalpha(1+2cos^2alpha-1+2cos^2alpha)}{cosalpha(cos2alpha+1-2sin^2alpha)}cdot2cos^22alpha=\ =frac{sinalphacdot4cos^2alpha}{cosalphacdot2cos2alpha}cdot2cos^22alpha=\ =4sinalphacosalphacos2alpha=2sin2alpha2cos2alpha=sin4alpha;$
3)
$ctg2alphafrac{2tgalpha}{1+tg^2alpha}=cos2alpha;\ frac{cos2alpha}{sin2alpha}cdotfrac{2frac{sinalpha}{cosalpha}}{1+(frac{sinalpha}{cosalpha})^2}=frac{cos2alpha}{sin2alpha}cdotfrac{2frac{sinalpha}{cosalpha}}{frac{cos^2alpha+sin^2alpha}{cos^2alpha}}=2cdotfrac{cos2alpha}{sin2alpha}cdotfrac{frac{sinalpha}{cosalpha}}{frac{1}{cos^2alpha}}=\ =2cdotfrac{cos2alpha}{sin2alpha}cdotfrac{sinalphacdotcos^2alpha}{cosalpha}=$
$=2cdotfrac{cos2alpha}{sin2alpha}cdotsinalphacdotcosalpha=frac{cos2alpha}{sin2alpha}cdot2sinalphacosalpha=frac{cos2alpha}{sin2alpha}cdotsin2alpha=cos2alpha;$