已知函数f(x)=2√2sin^2 (π/4+x)-cos2x,设三角形ABC的最小内角为角A满足f(A)=2...

题目

已知函数f(x)=2√2sin^2 (π/4+x)-cos2x,设三角形ABC的最小内角为角A满足f(A)=2...
已知函数f(x)=2√2sin^2 (π/4+x)-cos2x,设三角形ABC的最小内角为角A满足f(A)=2根3(1)求角A的大小

答案

由半角公式
sin^2 x = 1/2 (1-cos2x)
2sin^2 (pi/4+x) = 1-cos(pi/2 +2x）= 1+sin2x
所以
f(x) = sqrt(3) + sqrt(3)sin2x - cos2x = sqrt(3) + 2 sin(2x-pi/6)
f(A) = 2sqrt(3)
2sin(2A-pi/6) = sqrt(3)
sin(2A-pi/6) = sqrt(3)/2
2A - pi/6 = pi/3
A = pi/4