已知函数f(x)=f'(π/4)cos(x)+sin(x),则f(π/4)的值
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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
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](/uploads/image/z/18510673-49-3.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Df%27%28%CF%80%2F4%29cos%28x%29%2Bsin%28x%29%2C%E5%88%99f%28%CF%80%2F4%29%E7%9A%84%E5%80%BC)
f'(π/4)=-f'(π/4)sin(π/4)+cosπ/4
即
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
即
f’(π/4)=√2/(√2+2)=√2(√2-2)/(√2+2)(√2-2)
=√2(√2-2)/[-2]
=(2-2√2)/[-2]
=√2-1.
即
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
即
f’(π/4)=√2/(√2+2)=√2(√2-2)/(√2+2)(√2-2)
=√2(√2-2)/[-2]
=(2-2√2)/[-2]
=√2-1.
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