2道高一三角题1.已知sina+sinb+sinc=0,cosa+cosb+cosc=0,则cos(a-b)的值是多少.

(1)
sina+sinb+sinc=0→sina+sinb=-sinc ①
同理有 cosa+cosb=- cosc ②
①的平方+②的平方→(sina+sinb)^2+(cosa+cosb)^2=1
→sina*sina+2*sina*sinb+sinb*sinb+cosa*cosa+2*cosa*cosb+cosb*cosb=1
→(sina*sina+cosa*cosa)+(sinb*sinb+cosb*cosb)+2*(cosa*cosb+sina*sinb)=1
→1+1+2cos(a-b)=1
→cos(a-b)= -(1/2)
(2)
sin(派/4-3X)cos(派/3-3X)-cos(派/6+3X)sin(派/4+3X)
=cos(派/2-派/4+3X)cos(派/3-3X)-sin()sin(派/3-派/6-3X) sin(派/4+3X)
= cos(派/4+3X)cos(派/3-3X)-sin()sin(派/3-3X) sin(派/4+3X)
=cos(派/4+3X+派/3-3X)
=cos(7派/12)