若f(x)=asin(x+π/4)+bsin(x-π/4)(ab≠0)是偶函数,则a、b满足的关系式?
网友回答
f(x)=a[sinxcosπ/4+cosxsinπ/4]+b[sinxcosπ/4-cosxsinπ/4]
=√2a/2[sinx+cosx]+√2b/2[sinx-cosx]
f(-x)=√2a/2[sin(-x)+cos(-x)]+√2b/2[sin(-x)-cos(-x)]
=√2a/2[-sinx+cosx]+√2b/2[-sinx-cosx]
=√2a/2[-sinx+cosx]-√2b/2[sinx+cosx]
=f(x)比较系数得:
√2a/2=-√2b/2
a=-b======以下答案可供参考======
供参考答案1:
因为f(x)为偶函数,f(-x)=f(x)
所以asin(x+π/4)+bsin(x-π/4)
=asin(π/4 -x)+bsin(-x-π/4)
=-asin(x-π/4)-bsin(x+π/4)由此可得a=-b