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设f(n)=cos^nα+sin^nα(n属于Z),求证;2f(6)-3f(4)+1=0
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设f(n)=cos^n α+sin^n α(n属于Z),求证;2f(6)-3f(4)+1=0
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答案和解析
f(n)=cos^n α+sin^n
f(6)=cos^6 α+sin^6α
=(cos^2α+sin^2α)(cos^4α-cos^2αsin^2α+sin^4α)
=cos^4α-cos^2αsin^2α+sin^4α
而f(4)=cos^4 α+sin^4α
所以
2f(6)-3f(4)+1
=2(cos^4α-cos^2αsin^2α+sin^4α)-3(cos^4 α+sin^4α)+1
=-(cos^4 α+2cos^2αsin^2α+sin^4α)+1
=-(cos^2α+sin^2α)²+1
=-1+1
=0
f(6)=cos^6 α+sin^6α
=(cos^2α+sin^2α)(cos^4α-cos^2αsin^2α+sin^4α)
=cos^4α-cos^2αsin^2α+sin^4α
而f(4)=cos^4 α+sin^4α
所以
2f(6)-3f(4)+1
=2(cos^4α-cos^2αsin^2α+sin^4α)-3(cos^4 α+sin^4α)+1
=-(cos^4 α+2cos^2αsin^2α+sin^4α)+1
=-(cos^2α+sin^2α)²+1
=-1+1
=0
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