f(x)=sin(x+a)+cos(x-a)是偶函数,x≠kπ,求f(2x-2/3π)
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f(x)=sin(x+a)+cos(x-a)是偶函数,x≠kπ,求f(2x-2/3π)
k属于z
k属于z
f(-x)=sin(-x+a)+cos(-x-a)
=-sin(x-a)+cos(x+a)
f(x)是偶函数所以f(x)=f(-x)
即-sin(x-a)+cos(x-a)=sin(x+a)+cos(x+a)
√2cos(x-a+π/4)=√2cos(x+a-π/4),x-a+π/4=2kπ+x+a-π/4,a=-kπ+π/4
f(x)=sin(x+a)+cos(x-a)
=-sin(x-kπ+π/4)+cos(x+kπ-π/4)
当k=2n时f(x)=-sin(x-2nπ+π/4)+cos(x+2nπ-π/4)=-sin(x+π/4)+cos(x-π/4)=-sin(x+π/4)+sin(x+π/4)=0
所以f(2x-2/3π)=0
当k=2n+1时f(x)=-sin(x-2nπ-3π/4)+cos(x+2nπ+3π/4)=sin(x+π/4)-cos(x+π/4)=√2sin(x+π/4-π/4)=√2sinx
所以f(2x-2/3π)=√2sin(2x-2/3π)=-√2sin(2x+π/3)
前面多谢你提醒啊!
=-sin(x-a)+cos(x+a)
f(x)是偶函数所以f(x)=f(-x)
即-sin(x-a)+cos(x-a)=sin(x+a)+cos(x+a)
√2cos(x-a+π/4)=√2cos(x+a-π/4),x-a+π/4=2kπ+x+a-π/4,a=-kπ+π/4
f(x)=sin(x+a)+cos(x-a)
=-sin(x-kπ+π/4)+cos(x+kπ-π/4)
当k=2n时f(x)=-sin(x-2nπ+π/4)+cos(x+2nπ-π/4)=-sin(x+π/4)+cos(x-π/4)=-sin(x+π/4)+sin(x+π/4)=0
所以f(2x-2/3π)=0
当k=2n+1时f(x)=-sin(x-2nπ-3π/4)+cos(x+2nπ+3π/4)=sin(x+π/4)-cos(x+π/4)=√2sin(x+π/4-π/4)=√2sinx
所以f(2x-2/3π)=√2sin(2x-2/3π)=-√2sin(2x+π/3)
前面多谢你提醒啊!
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