高一数学。 手写过程
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分你为奇数和偶数讨论
(1)当n为偶数,即n=2k,(k∈Z)时,
f(x)= cos2(2kπ+x)•sin2(2kπ-x) cos2[(2×2k+1)π-x] = cos2x•sin2(-x) cos2(π-x) = cos2x•(-sinx)2 (-cosx)2 =sin2x,(n∈Z)
当n为奇数,即n=2k+1,(k∈Z)时f(x)= cos2[(2k+1)π+x]•sin2[(2k+1)π-x] cos2{[2×(2k+1)+1]π-x} = cos2[2kπ+(π+x)]•sin2[2kπ+(π-x)] cos2[2×(2k+1)π+(π-x)] = cos2(π+x)•sin2(π-x) cos2(π-x) = (-cosx)2•sin2x (-cosx)2 =sin2x,(n∈Z)
∴f(x)=sin2x;
(2)由(1)得f( π/ 2010 )+f( 502π/ 1005 )=sin^2 π /2010 +sin^2 1004π/ 2010
=sin^2 π/ 2010 +sin^2( π/ 2 - π/ 2010 )=sin^2 π/ 2010 +cos^2( π/ 2010 )=1
(1)当n为偶数,即n=2k,(k∈Z)时,
f(x)= cos2(2kπ+x)•sin2(2kπ-x) cos2[(2×2k+1)π-x] = cos2x•sin2(-x) cos2(π-x) = cos2x•(-sinx)2 (-cosx)2 =sin2x,(n∈Z)
当n为奇数,即n=2k+1,(k∈Z)时f(x)= cos2[(2k+1)π+x]•sin2[(2k+1)π-x] cos2{[2×(2k+1)+1]π-x} = cos2[2kπ+(π+x)]•sin2[2kπ+(π-x)] cos2[2×(2k+1)π+(π-x)] = cos2(π+x)•sin2(π-x) cos2(π-x) = (-cosx)2•sin2x (-cosx)2 =sin2x,(n∈Z)
∴f(x)=sin2x;
(2)由(1)得f( π/ 2010 )+f( 502π/ 1005 )=sin^2 π /2010 +sin^2 1004π/ 2010
=sin^2 π/ 2010 +sin^2( π/ 2 - π/ 2010 )=sin^2 π/ 2010 +cos^2( π/ 2010 )=1
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