因为 f(sin x)=sin nx,
所以 f[ sin (x +pi/2) ]=sin [ n (x +pi/2) ].
即 f(cos x)=sin (nx +n/2 *pi)
=sin (nx) cos (n/2 *pi) +cos (nx) sin (n/2 *pi).
要使 f(cos x) =cos nx ,比较系数得
cos (n/2 *pi)=0,
sin (n/2 *pi)=1.
所以 n/2 *pi =pi/2 +2k *pi,k为整数.
所以 n=4k+1,k为整数.
即 当n=4k+1,k为整数时,有
f(cos x) =cos nx.
= = = = = = = = =
待定系数法.用sin (x +pi/2) =cos x 比较简单,用sin (pi/2 -x) =cos x 就麻烦一点.
= = = = = = = = =
解法2：因为 f(sin x) =sin nx,
所以 f[ sin (x +pi/2) ] =sin [ n (x +pi/2) ].
即 f(cos x) =sin (nx +n/2 *pi).
i) 若 n=4k,k为整数,
则 f(cos x) =sin (nx +2k *pi)
=sin (nx).
ii) 若 n=4k+1,k为整数,
则 f(cos x) =sin (nx +2k *pi +pi/2)
=sin (nx +pi/2)
=cos (nx).
iii) 若 n=4k+2,k为整数,
则 f(cos x) =sin (nx +2k *pi +pi)
=sin (nx +pi)
= -sin (nx).
iv) 若 n=4k+3,k为整数,
则 f(cos x) =sin (nx +2k *pi +3pi/2)
= sin (nx +3pi/2)
= -cos (nx).
综上,当且仅当 n=4k+1 时,
f(cos x) =cos nx.
= = = = = = = = =
以上计算可能有误,你最后检查一下.
分类讨论.
这种方法,诱导公式要很熟才行,否则容易出错.