Question

已知tanαtanβ=根号3/3,求证(2-cos2α)(2-cos2β)=3

Answer 1

(2sin²a+1)(2sin²b+1)
=4(sinasinb)²+2sin²a+2sin²b+1
sin²asin²b/cos²acos²b=1/3
所以cos²acos²b=3sin²asin²b=(1-sin²a)(1-sin²b)
展开：2sin²asin²b+sin²a+sin²b=1
所以原式=2×1+1
=3

Answer 2

tanatanb=√3/3

由万能公式
(2-cos2a)(2-cos2b)
=[2-(1-tan²a)/(1+tan²a)][2-(1-tan²b)/(1+tan²b)]
=(1+3tan²a)/(1+tan²a)*(1+3tan²b)/(1+tan²b)
=(1+3tan²a+3tan²b+9tan²atan²b)/(1+tan²a+tan²b+tan²atan²b)
=(1+3tan²a+3tan²b+3)/(1+tan²a+tan²b+1/3)
=3(4/3+tan²a+tan²b)/(4/3+tan²a+tan²b)
=3
命题得证