将函数f(x)=sinx展开成(x-π/4)的幂级数这样做:f(x)=sinx=sin(x-π/4+π/4)=根号2/2(sin(x-π/4)+cos(x-π/4))再讲x-π/4直接代入sinx和cosx的麦克劳林展开式中行不?

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将函数f(x)=sinx展开成(x-π/4)的幂级数这样做:f(x)=sinx=sin(x-π/4+π/4)=根号2/2(sin(x-π/4)+cos(x-π/4))再讲x-π/4直接代入sinx和cosx的麦克劳林展开式中行不?
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将函数f(x)=sinx展开成(x-π/4)的幂级数这样做:f(x)=sinx=sin(x-π/4+π/4)=根号2/2(sin(x-π/4)+cos(x-π/4))再讲x-π/4直接代入sinx和cosx的麦克劳林展开式中行不?
将函数f(x)=sinx展开成(x-π/4)的幂级数
这样做:f(x)=sinx=sin(x-π/4+π/4)=根号2/2(sin(x-π/4)+cos(x-π/4))
再讲x-π/4直接代入sinx和cosx的麦克劳林展开式中行不?

将函数f(x)=sinx展开成(x-π/4)的幂级数这样做:f(x)=sinx=sin(x-π/4+π/4)=根号2/2(sin(x-π/4)+cos(x-π/4))再讲x-π/4直接代入sinx和cosx的麦克劳林展开式中行不?
可以的.因为sinx和cosx的麦克劳林公式对所有实数都成立.