Question

求这几道题详细解答过程，谢谢

Answer 1

11、
sinA+sin3A+sin5A=sin3A+(sinA+sin5A)=sin3A+2sin3Acos2A=sin3A(1+2cos2A)=a
cosA+cos3A+cos5A=cos3A+(cosA+cos5A)=cos3A+2cos3Acos2A=cos3A(1+2cos2A)=b
所以，a^2+b^2=sin^2 3A*(1+2cos2A)^2+cos^2 3A*(1+2cos2A)^2
=(1+2cos2A)^2*(sin^2 3A+cos^2 3A)
=(1+2cos2A)^2

12、
cos^2 x+cos^2 (x+α)-2cosxcosαcos(x+α)
=cos^2 x+cos(x+α)[cos(x+α)-2cosxcosα]
=cos^2 x+cos(x+α)(cosxcosα-sinxsinα-2cosxcosα)
=cos^2 x-cos(x+α)(cosxcosα+sinxsinα)
=cos^2 x-cos(x+α)cos(x-α)
=cos^2 x-(1/2)(cos2x+cos2α)
=cos^2 x-(1/2)[(2cos^2 x-1)+(1-2sin^2 α)]
=cos^2 x-cos^2 x+sin^2 α
=sin^2 α

13、
y=cos3xcosx=(1/2)(cos4x+cos2x)=(1/2)[(2cos^2 2x-1)+cos2x]
=cos^2 2x+(1/2)cos2x-(1/2)
令cos2x=t∈[-1,1]
则，y=t^2+(1/2)t-(1/2)=[t^2+(1/2)t+(1/16)]-(1/16)-(1/2)
=[t+(1/4)]^2-(9/16)
所以，当t=-1/4时，y有最小值=-9/16
当t=1时，y有最大值=1