1.因为(3π/4+b)-(π/4-a)=b+a+π/2所以cos[（a+b）+π/2]=cos(a+b)*cosπ/2-sin(a+b)*sinπ/2=-sin(a+b)所以sin(a+b)=-cos[（a+b）+π/2]=-cos[(3π/4+b)-(π/4-a)] =-[cos(3π/4+b)*cos(π/4-a)+sin(3π/4+b...