已知函数f(x)=f'(π/4)cos(x)+sin(x),则f(π/4)的值我看了答案,对下面的这一步不理解f'(π/4)=-f'(π/4)sin(π/4)+cosπ/4是如何转化为f'(π/4)=√2 -1

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已知函数f(x)=f'(π/4)cos(x)+sin(x),则f(π/4)的值我看了答案,对下面的这一步不理解f'(π/4)=-f'(π/4)sin(π/4)+cosπ/4是如何转化为f'(π/4)=√2 -1
已知函数f(x)=f'(π/4)cos(x)+sin(x),则f(π/4)的值
我看了答案,对下面的这一步不理解
f'(π/4)=-f'(π/4)sin(π/4)+cosπ/4是如何转化为f'(π/4)=√2 -1

已知函数f(x)=f'(π/4)cos(x)+sin(x),则f(π/4)的值我看了答案,对下面的这一步不理解f'(π/4)=-f'(π/4)sin(π/4)+cosπ/4是如何转化为f'(π/4)=√2 -1
f'(π/4)=-f'(π/4)sin(π/4)+cosπ/4

f'(π/4)+f'(π/4)sin(π/4)=cosπ/4
f'(π/4)+f'(π/4)√2/2=√2/2
2f'(π/4)+f'(π/4)√2=√2
(2+√2)f‘(π/4)=√2

f’(π/4)=√2/(√2+2)=√2(√2-2)/(√2+2)(√2-2)
=√2(√2-2)/[-2]
=(2-2√2)/[-2]
=√2-1.

f(x)=f'(π/4)cos(x)+sin(x)
求导得到f‘(x)=-f'(π/4)sinx+cosx
再令x=π/4
得到f'(π/4)=-f'(π/4)sin(π/4)+cosπ/4=-√2/2f'(π/4)+√2/2
解出f'(π/4)=√2 -1