Question

помогите 1 , 2 задания

Answer 1

$1.;a);frac{sin2a}{sin^2a}=frac{2sin acos a}{sin^2a}=frac{2cos a}{sin a}=2ctg;a\b);frac{sin2a}{2sin a}=frac{2sin acos a}{2sin a}=cos a\c);sin^2a+cos2a=sin^2a+cos^2a-sin^2a=cos^2a\d);frac{cos2a}{cos a-sin a}-sin a=frac{cos^2a-sin^2a}{cos a-sin a}-sin a=\=frac{(cos a-sin a)(cos a+sin a)}{cos a-sin a}-sin a=cos a+sin a-sin a=cos a$

$2.;sin a=frac5{13}\sin^2a+cos^2a=1\cos a=sqrt{1-sin^2a}=sqrt{1-left(frac5{13}right)^2}=sqrt{1-frac{25}{169}}=sqrt{frac{144}{169}}=pmfrac{12}{13}\fracpi2<a<piRightarrowcos a=-frac{12}{13}\sin2a=2sin acos a=2cdotfrac5{13}cdot(-frac{12}{13})=-frac{120}{169}\cos2a=cos^2a-sin^2a=left(-frac{12}{13}right)^2-left(frac{15}{13}right)^2=frac{144}{169}-frac{25}{169}=frac{119}{169}$