已知函数f(x)=cos²x+sinxcosx.(1)求函数f(x)的最大值;(2)在△ABC中,AB=AC=3,角A满足f(A/2+π/8)=1,求面积.
问题描述:
已知函数f(x)=cos²x+sinxcosx.(1)求函数f(x)的最大值;
(2)在△ABC中,AB=AC=3,角A满足f(A/2+π/8)=1,求面积.
答
(1)
f(x)=(1+cos2x)/2+(1/2)sin2x
=(√2/2)sin(2x+π/4)+1/2.
∴sin(2x+π/4)=1时,
y|min=(1+√2)/2.
(2)
f(A/2+π/8)=1,
∴(√2/2)sin[2(A/2+π/8)+π/8]+1/2=1
→sin(A+3π/8)=√2/2
→A+3π/8=3π/4
→A=3π/8.
∴三角形面积S=(1/2)·3·3·sin(3π/8)