題目：
三角形ABC中角A.B.C所對的邊分別為a,b,c且滿足csinA=acosC
⑴求角C大小?⑵求√3sinA-cos(B+C)的最大值,并求取得最大值時角A B的大小.
答案：
(1)、在△ABC中
∵csinA=acosC
∴a/sinA=c/cosC
又∵根據(jù)正弦定理：a/sinA=c/sinC
∴sinC=cosC
∴∠C=45°

(2)、√3sinA-cos(B+C)
=√3sin(B+C)-cos(B+C)
=2[√3/2*sin(B+C)-1/2*cos(B+C)]
=2sin(B+C-π/6)
=2sin(B+π/4-π/6)
=2sin(B+π/12)
∵∠C=π/4
∴∠A+∠B=3π/4
即0<∠B＜3π/4
∴π/12<∠B+π/12<5π/6
∴當∠B+π/12=π/2時,sin(B+π/12)取得最大值1
即當∠B=5π/12時,√3sinA-cos(B+C)取得最大值2
此時∠A=π-∠B-∠C=π/3