Răspuns:
sin(a+b)=sin(a)cos(b)+cos(a)sin(b)
sin(a) + cos(b)= 1 ⇔(sin(a) + cos(b))²= 1² ⇔
sin(a)²+cos(b)²+2sin(a)cos(b)= 1 (1)
cos(a) + sin(b)= -1/2 ⇔[cos(a) + sin(b)]²= (-1/2)² ⇔
cos(a)² + sin(b)²+2cos(a)sin(b)=1/4 (2)
adunam relatiile (1) si (2) ⇒sin(a)²+cos(b)²+2sin(a)cos(b)+cos(a)² + sin(b)²+2cos(a)sin(b)=1+1/4 ⇔
[sin(a)²+cos(a)²]+[cos(b)²+sin(b)²] + 2[sin(a)cos(b)+cos(a)sin(b)]=1+1/4 ⇔
1+1+2sin(a+b)=1+1/4 ⇔2sin(a+b)=1/4-1 ⇔2sin(a+b) = -3/4
sin(a+b) = -3/8
