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)
答
tan2A=tan((A+B)+(A-B))=(tan(A+B)+tan(A-B))/(1-tan(A+B)tan(A-B))=(3+5)/(1-3*5)=-4/7tan2B=tan((A+B)-(A-B))=(tan(A+B)-tan(A-B))/(1+tan(A+B)tan(A-B))=(3-5)/(1+15)=-1/8