Question

Решите тождества:

1)sin(5pi/4 + a) = - sin (3pi/4 - a)

2) cos (a - 2pi/3)= cos (a + 4pi/3)

 

Answer 1

1)$sin(frac{5pi}{4}+a)=-sin(frac{3pi}{4}-a)\sinfrac{5pi}{4}*cosa+sina*cosfrac{5pi}{4}=-(sinfrac{3pi}{4}*cosa-sina*cosfrac{3pi}{4})\sin(pi+frac{pi}{4})*cosa+sina*cos(pi+frac{pi}{4})=-sin(pi-frac{pi}{4})*cosa+sina*cos(pi-frac{pi}{4})\-sinfrac{pi}{4}*cosa+sina*-cosfrac{pi}{4}=-sinfrac{pi}{4}*cosa+sina*-cosfrac{pi}{4}$

Доказано. 

 

$cos(a-frac{2pi}{3})=cos(a+frac{4pi}{3})\cosa*cosfrac{2pi}{3}+sina*sinfrac{2pi}{3}=cosa*cosfrac{4pi}{3}-sina*sinfrac{4pi}{3}\cosa*cos(pi-frac{pi}{3})+sina*sin(pi-frac{pi}{3})=cosa*cos(pi+frac{pi}{3})-sina*sin(pi+frac{pi}{3})\cosa*-cosfrac{pi}{3}+sina*sinfrac{pi}{3}=cosa*-cosfrac{pi}{3}-(-sinfrac{pi}{3})*sina$

Доказано.