Question

(sinx)^4+(cosx)^4=?

Answer 1

先设sinxcosx=t
(sinx)^4 (cosx)^4
=[(sinx)^3 (cosx)^3](sinx cosx)-(sinx)^3*cosx-(cosx)^3*sinx
=sinx cosx-sinxcosx[(sinx)^2 (cosx)^2]
=sinx cosx-sinxcosx
=sinxcosx-t
又(sinx)^4 (cosx)^4=[(sinx)^2 (cosx)^2]^2-2(sinx)^2(cosx)^2
=1-2t^2
所以sinx cosx=1 t-2t^2
所以利用条件的等式有:
(sinx)^3 (cosx)^3=1=(sinx cosx)[(sinx)^2-sinxcosx (cosx)^2]
=(1 t-2t^2)(1-t)
=1-3t^2 2t^3
所以t^2(2t-3)=0
那么t=0或者t=3/2(t=1/2sin2x范围是[-1/2,1/2],所以这个值舍去)
那么sinxcosx=0且(sinx)^3 (cosx)^3=1
所以sinx和cosx中有一个为0,有一个为1
所以就有
sinx cosx=(sinx)^4 (cosx)^4=1

Answer 2

sin�6�6x + cos�6�6x
= (sin�6�6x + cos�6�6x + 2sin�0�5xcos�0�5x) - 2sin�0�5xcos�0�5x
= (sin�0�5x+cos�0�5x)�0�5 - 2(sinxcosx)�0�5
= (1)�0�5 - 2(1/2 * sin2x)�0�5
= 1 - (1/2)sin�0�5(2x)
= 1 - (1/2)(1/2)(1 - cos4x)
= 1 - (1/4)(1 - cos4x)
= 1 - 1/4 + (1/4)cos4x
= (1/4)(3 + cos4x)