Question

sin77°-sin 55-sin 103°+sin65°=sin5° доказать тождества​

Answer 1

Объяснение:

$\displaystyle\\sin77^0-sin55^0-sin103^0+sin65^0=(sin65^0-sin55^0)-(sin103^0-sin77^0)=\\\\\\=2*sin\frac{65^0-55^0}{2}* cos\frac{65^0+55^0}{2} -2*sin\frac{103^0-77^0}{2}*cos\frac{103^0+77^0}{2}=\\\\\\=2*sin5^0 *cos60^0-2*sin13^0*cos90^0=2*sin5^0*\frac{1}{2} -2*sin13^0*0=sin5^0.$