函数f(x)=2cos2x+sin2x的最小值是().
参考答案:
-√2+1
解析:
f(x)=1+cos2x+sin2x
=√2(√2/2*sin2x+√2/2cos2x)+1
=√2(sin2xcosπ/4+cos2xsinπ/4)+1
=√2sin(2x+π/4)+1
函数f(x)=2cos2x+sin2x的最小值是().
参考答案:
-√2+1
解析:
f(x)=1+cos2x+sin2x
=√2(√2/2*sin2x+√2/2cos2x)+1
=√2(sin2xcosπ/4+cos2xsinπ/4)+1
=√2sin(2x+π/4)+1