f(x)=asin(x+π/4)+bsin(x-π/4) (ab不为0)是偶函数,有序实数对(a,b)可以是?

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f(x)=asin(x+π/4)+bsin(x-π/4) (ab不为0)是偶函数,有序实数对(a,b)可以是?

f(x)=f(-x)
f(x)=asin(x+π/4)+bsin(x-π/4)
f(-x)=asin(-x+π/4)+bsin(-x-π/4)
=-bsin(x+π/4)-asin(x-π/4)
所以a=-b
我认为只要满足a=-b的都可以

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-...

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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

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