sin^2(160)+cos^2(310)+sin(160)*cos(310) 怎么算?

sin^2(160)+cos^2(310)+sin(160)*cos(310) 怎么算?
sin^2(160)+cos^2(310)+sin(160)*cos(310) 怎么算?

sin^2(160)+cos^2(310)+sin(160)*cos(310) 怎么算?
(常规方法我不会,不过感觉这一类题好像都这样做)
由余弦定理：cosC=(a²+b²-c²)/2ab 和正弦定理：a/sinA=b/sinB=c/sinC
得sin²A+sin²B-sin²C=2sinAsinBcosC.
即sin²C=sin²A+sin²B - 2sinAsinBcosC.
sin² (160°)=sin²(180°-160°)=sin²20°
cos² (310°)=cos² (360°-310°)=cos²50°=sin²40°
sin(160°)cos(310°)=sin(20°)cos(50°)=sin(20°)sin(40°)
为了凑上面的形式,令A=20°,B=40°,C=180°- A - B=120°.
则sin(20°)sin(40°)= -2sin(20°)sin(40°)cos(120°).
由上面可知sin²(120°)=sin²20°+sin²40° - 2sin(20°)sin(40°)cos(120°)=3/4
此即所求式子的值.
如果用降次与和差化积化简的话也可以：
(接上面的)sin²20°=(1-cos40°)/2=1/2-½sin50°
sin²40°=(1-cos80°)/2=1/2-½sin10°
sin(20°)cos(50°)=½(sin70°+sin(-30°))=½sin70°- 1/4
相加得 -½(sin10°+sin50° -sin70°)+3/4.
而sin10°+sin50°=2sin30°cos20°=cos20°=sin70°.
∴ -½(sin10°+sin50° -sin70°)=0,即原式的值为3/4.