求证:3+tan1tan2+tan2tan3=tan3/tan1
1个回答
2013-05-09
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设 tan1=t
则由二倍、三倍角公式
tan2=2t/(1-t^2) tan3=(3t-t^3)/(1-3t^2)
3+tan1tan2+tan2tan3
=3+2t^2/(1-t^2)+2t(3t-t^3)/(1-t^2)(1-3t^2)
=(t^4-4t^2+3)/(t^2-1)(1-3t^2)
=(3-t^2)/(1-3t^2)
=tan3/tan1
则由二倍、三倍角公式
tan2=2t/(1-t^2) tan3=(3t-t^3)/(1-3t^2)
3+tan1tan2+tan2tan3
=3+2t^2/(1-t^2)+2t(3t-t^3)/(1-t^2)(1-3t^2)
=(t^4-4t^2+3)/(t^2-1)(1-3t^2)
=(3-t^2)/(1-3t^2)
=tan3/tan1
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