Question

请问第二问怎么做

Answer 1

√2cosA=2cosAcosB=cos(A+B)+cos(A-B)=-cosC+cos(A-B)

∴原式=cos(A-B)=cos(A-π/4)
∵A∈(0,3π/4),∴A-π/4∈(-π/4,π/2)
在该区间上馀弦能取到最大值1,此时A=π/4

Answer 2

C=π - A - B=π - A - π/4
=3π/4 - A
∴原式=√2cosA + cos(3π/4 - A)
=√2cosA + cos(3π/4)cosA + sin(3π/4)sinA
=√2cosA - (√2/2)cosA + (√2/2)sinA
=(√2/2)sinA + (√2/2)cosA
=sin(A + π/4)
∴最大值是1