已知函数f(x)=asin(πx+α)+bcos(πx+β),其中a,b,α,β都是非零实数,且满足f(2009)=2,则f(2010)=?
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![已知函数f(x)=asin(πx+α)+bcos(πx+β),其中a,b,α,β都是非零实数,且满足f(2009)=2,则f(2010)=?](/uploads/image/z/3757030-70-0.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dasin%28%CF%80x%2B%CE%B1%29%2Bbcos%28%CF%80x%2B%CE%B2%29%2C%E5%85%B6%E4%B8%ADa%2Cb%2C%CE%B1%2C%CE%B2%E9%83%BD%E6%98%AF%E9%9D%9E%E9%9B%B6%E5%AE%9E%E6%95%B0%2C%E4%B8%94%E6%BB%A1%E8%B6%B3f%282009%29%3D2%2C%E5%88%99f%282010%29%3D%3F)
已知函数f(x)=asin(πx+α)+bcos(πx+β),其中a,b,α,β都是非零实数,且满足f(2009)=2,则f(2010)=?
已知函数f(x)=asin(πx+α)+bcos(πx+β),其中a,b,α,β都是非零实数,且满足f(2009)=2,则f(2010)=?
已知函数f(x)=asin(πx+α)+bcos(πx+β),其中a,b,α,β都是非零实数,且满足f(2009)=2,则f(2010)=?
∵sin(A+π)=-sinA cos(A+π)=-cosA
知道了三角函数的诱导公式下面就好办了
f(2010)=asin(2010π+α)+bcos(2010π+β)
=asin(2009π+α+π)+bcos(2009π+β+π)
=-asin(2009π+α)-bcos(2009π+β)
=-f(2009)=-2
f(x)=asin(πx+α)+bcos(πx+β)
∵f(2009)=asin(2009π+α)+bcos(2009π+β)
=asin(π+α)+bcos(π+β)
=-(asinα+bcosβ)
=5
∴asinα+bcosβ=-5
∴f(2014)=asin(2014π+α)+bcos(2014π+β)
=asinα+bcosβ
=-5
满意请采纳。
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