Question

если sinA=4/5,и cosB=12/13 и A и В углы острые,то найдите 13sin(A+B).

Answer 1

$sin alpha=frac{4}{5}\ \ cos^2 alpha+sin^2 alpha=1\\ cos^2 alpha= 1-sin^2 alpha\ \ cos alpha=sqrt{1-sin^2 alpha}=sqrt{1-(frac{4}{5})^2}=sqrt{1-frac{16}{25}} =sqrt{frac{25-16}{25}}=sqrt{frac{9}{25}}=frac{3}{5} \\ cos beta=frac{12}{13}\ \ cos^2beta+sin^2 beta=1\\ cos^2beta= 1-sin^2 beta\ \ sin beta=sqrt{1-cos^2 beta}=sqrt{1-(frac{12}{13})^2}=sqrt{1-frac{144}{169}} =sqrt{frac{169-144}{169}}=\\=sqrt{frac{25}{169}}=frac{5}{13}$

$13sin(alpha +beta )=13(sinalpha cdotcosbeta +cosalpha cdotsinbeta )=13(frac{4}{5} cdotfrac{12}{13}+frac{3}{5} cdotfrac{5}{13})=\ \ =frac{48}{5}+3=9,6+3=12,6$

Ответ: $12,6$