Question

这个高二数学题求解

Answer 1

(1) tan(A/2)+cot(A/2)=1/(sinA/2*cosA/2)=2/sinA=10/3,=>sinA=3/5;
∵sinC=sin(A+B)>0 => cosA>-1/4 ∴cosA=4/5;cosB=5/13 sinB=12/13; cos(A-B)=cosAcosB+sinAsinB=56/65;
(2) ∵cos(A-B)>0 cosA>cosB ∴A<B 且 -90<A-B<0 => -45<(A-B)/2<0
∴cos[(A-B)/2]>0 ∴cos[(A-B)/2]=根号下[1+cos(A-B)/2]=(11*根号130)/130;
第二题:
(1) y=cosC[cos(A-B)-cosC]+2=cosC[cos(A-B)+cos(A+B)]+2=2cosCcosAcosB+2;观察可知任意交换A,B,C的位置,y的值不变.
(2) y=-[cos^2 C-cosCcos(A-B)]+2
=-{[cosC-1/2cos(A-B)]^2-1/4cos^2 (A-B)}+2
=-[cosC-1/2cos(A-B)]^2+1/4cos^2 (A-B)+2
<=1/4cos^2 (A-B)+2
=1/4[1-sin^2 (A-B)] +2
=-1/4sin^2 (A-B) +9/4
<=9/4
当且仅当 cosC-1/2cos(A-B)=0 sin^2 (A-B)=0 即 三角形为等边时取等 最大值为9/4;