已知f(x)=asin(πx+a)+bcos(π+b)+4,且f(2010)=3,求f(2011)的值
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![已知f(x)=asin(πx+a)+bcos(π+b)+4,且f(2010)=3,求f(2011)的值](/uploads/image/z/2554932-12-2.jpg?t=%E5%B7%B2%E7%9F%A5f%28x%29%3Dasin%28%CF%80x%2Ba%29%2Bbcos%28%CF%80%2Bb%29%2B4%2C%E4%B8%94f%282010%29%3D3%2C%E6%B1%82f%282011%29%E7%9A%84%E5%80%BC)
已知f(x)=asin(πx+a)+bcos(π+b)+4,且f(2010)=3,求f(2011)的值
已知f(x)=asin(πx+a)+bcos(π+b)+4,且f(2010)=3,求f(2011)的值
已知f(x)=asin(πx+a)+bcos(π+b)+4,且f(2010)=3,求f(2011)的值
式子有误吧,应是f(x)=asin(πx+a)+bcos(πx+b)+4
f(2010)=asin(2010π+a)+bcos(2010π+b)+4
f(2011)=asin(2011π+a)+bcos(2011π+b)+4
=asin(π+2010π+a)+bcos(π+2010π+b)+4
=-[asin(2010π+a)+bcos(2010π+b)+4]+8
=-f(2010)+8
=-3+8=5
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