已知f(x)= 根号2 * sin(2x + pai/4 ) + 根号2 ,求F(x)=f(x)-f(x - pai/4
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已知f(x)= 根号2 * sin(2x + pai/4 ) + 根号2 ,求F(x)=f(x)-f(x - pai/4)的最大值、最小值
f(x)=√2*sin(2x+π/4)+√2
F(x)
=f(x)-f(x-π/4)
=√2*sin(2x+π/4)+√2-√2*sin(2(x-π/4)+π/4)-√2
=√2*sin(2x+π/4)-√2*sin(2x-π/4)
=√2*sin2x*√2/2+√2*cos2x*√2/2-√2*sin2x*√2/2+√2*cos2x*√2/2
=sin2x+cos2x-sin2x+cos2x
=2cos2x
∵-1
F(x)
=f(x)-f(x-π/4)
=√2*sin(2x+π/4)+√2-√2*sin(2(x-π/4)+π/4)-√2
=√2*sin(2x+π/4)-√2*sin(2x-π/4)
=√2*sin2x*√2/2+√2*cos2x*√2/2-√2*sin2x*√2/2+√2*cos2x*√2/2
=sin2x+cos2x-sin2x+cos2x
=2cos2x
∵-1
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