f(x)=2cos^2 x+2sin x cos x 求f(x)的最大值和最小值

利用2倍角公式
2sinxcosx=sin2x
cos2x=2(cosx)^2-1
cos2x+1=2(cosx)^2
f(x)=cos2x+1+sin2x
=√2sin(2x+π/4)+1
最大值√2+1
最小值-√2+1

f(x)=2cos^2 x+2sin x cos x=2cos^2 x-1+2sin x cos x+1
=cos2x+sin2x+1=√2sin(2x+pai/4)+1
所以f(x)max=√2+1
f(x)min=√2-1

f(x)=cos2x+1+sin2x=根号2＊cosy+1
max=根号2＋1
min=－根号2＋1

f(x)=1+cos2x+sin2x
=2^(1/2)cos(2x-3.14.../4)+1
max=1+2^(1/2)
min =1-2^(1/2)