设▲ABC的三个内角A、B、C所对的边a、b、c、且满足csinA=acosC.若根号3sinA-cos(B+π/4)的
1个回答

csinA=acosC => a/c = sinA/cosC

由正弦定理 a/c = sinA/sinC

∴ sinC =cosC => ∠C = π/4

∴ ∠A + ∠B = 3π/4 ==> ∠B = 3π/4 - ∠A

3sinA - cos(B+π/4)

= 3sinA - cos( 3π/4 - A +π/4)

= 3sinA + cosA

= √10*sin(A+θ)

其中 sinθ = √10/10;tanθ = 1/3

∵ 0< tanθ < √3/3

∴ 0 < θ < π/6

∠A 的取值范围是 (0,3π/4 )

因此 3sinA - cos(B+π/4) = √10*sin(A+θ) 的最大值为√10;

无法得出 A为直角的结论,只要 C= π/4,等式就成立;

A 可在(0,3π/4 )上任意取值.