tan2a
=tan[(a+b)+(a-b)]
=[tan(a+b)+tan(a-b)]/[1-tan(a+b)tan(a-b)]
=(2/5+1/4)/(1-2/5*1/4)
=(13/20)/(18/20)
=13/18
sin2A=±13/√(13²+18²)
=±13/√493
=(±13/493)√493
tan(a+b)=25,是2/5还是5/2呀
tan2a
=tan[(a+b)+(a-b)]
=[tan(a+b)+tan(a-b)]/[1-tan(a+b)tan(a-b)]
=(2/5+1/4)/(1-2/5*1/4)
=(13/20)/(18/20)
=13/18
sin2A=±13/√(13²+18²)
=±13/√493
=(±13/493)√493
tan(a+b)=25,是2/5还是5/2呀