Question

тригонометрия решить 1,2,3

Answer 1

1)
$sin^2a+cos^2a=1\cosa=pmsqrt{1-sin^2a}$
Угол второй четверти⇒косинус отрицательный.
$cosa=-sqrt{frac{16}{16}-frac{9}{16}}=-frac{sqrt{7}}{4}$

$sin(a+frac{pi}{4})=sina*cosfrac{pi}{4}+cosa*sinfrac{pi}{4}=frac{3}{4}*frac{sqrt{2}}{2}-frac{sqrt{7}}{4}*frac{sqrt{2}}{2}=frac{3sqrt{2}-sqrt{14}}{8}$



2)
$sinfrac{4pi}{5}*cosfrac{pi}{6}-sinfrac{pi}{6}*cosfrac{pi}{5}=sin(pi-frac{pi}{5})*cosfrac{pi}{6}-sinfrac{pi}{6}*cosfrac{pi}{5}=\=sinfrac{pi}{5}*cosfrac{pi}{6}-sinfrac{pi}{6}*cosfrac{pi}{5}=sin(frac{pi}{5}-frac{pi}{6})=sinfrac{pi}{30}$



3)
$cosa+cos(a-frac{pi}{3})=2cosfrac{a+(a-frac{pi}{3})}{2}*cosfrac{a-(a-frac{pi}{3})}{2}=\=2cos(a-frac{pi}{6})*cosfrac{pi}{6}$



4)
$sin(a+frac{3pi}{4})*cos(a-frac{pi}{4})=\=frac{1}{2}(sin(a+frac{3pi}{4}-(a-frac{pi}{4}))+sin(a+frac{3pi}{4}+(a-frac{pi}{4})))=frac{sinpi}{2}+frac{sin(2a+frac{pi}{2})}{2}$