tan(A+B)=3,tan(A-B)=5,求tan2A,tan2B.(A,B是角)

问题描述:

tan(A+B)=3,tan(A-B)=5,求tan2A,tan2B.(A,B是角)

tan2A=tan[(A+B)+(A-B)]
=[tan(A+B)+tan(A-B)]/[1-tan(A+B)tan(A-B)]
=(3+5)/(1-3×5)
=-4/7
tan2B=tan[(A+B)-(A-B)]
=[tan(A+B)-tan(A-B)]/[1+tan(A+B)tan(A-B)]
=(3-5)/(1+3×5)
=-1/8