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令F(x)=f(x)sinx,則F(x)在[0,π]上連續,在(0,π)內可導,且F(0)=F(π)=0,由羅爾定理,存在一點ε∈(0,π),使得F'(ε)=0,即f'(ε)sinε+f(ε)cosε=0
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