Question

已知tanatanb=根号3/3,求证：（2-cos2a)(2-cos2b)=3

Answer 1

【1】由“倍角公式”可得：cos2A=cos²A-sin²A,且2=2(sin²A+cos²A) ,∴2-cos2A=2sin²A+2cos²A-cos²A+sin²A=3sin²A+cos²A=(3sin²A+cos²A)/(sin²A+cos²A)=(3tan²A+1)/(tan²A+1).即：2-cos2A=(3tan²A+1)/(tan²A+1).同理可得：2-cos2B=(3tan²B+1)/(tan²B+1).【2】易知,tan²Atan²B=1/3.∴左边=[(3tan²A+1)(3tan²B+1)]/[(tan²A+1)(tan²B+1)]=[9tan²Atan²B+3(tan²A+tan²B)+1]/[tan²Atan²B+(tan²A+tan²B)+1]=[4+3(tan²A+tan²B)]/[(1/3)+(tan²A+tan²B)+1]=3[4+3(tan²A+tan²B)]/[4+3(tan²A+tan²B)]=3=右边.