cos(A+B)cos(A-B)=1/4 求cosA^2+cosB^2

这一题属于积化和差的类型，
cos(A+B)cos(A-B)=1/2 * {cos[(A+B)+(A-B)] + cos[(A+B)-(A-B)]}
=1/2(cos2A+cos2B)=1/4 故(cos2A+cos2B)=1/2
而cosA^2+cosB^2 =（1+cos2A）/2 +(1+cos2B)/2=1+(cos2A+cos2B)/2
=1+1/2*1/2=5/4

cos(A+B)cos(A-B)
=(cosAcosB-sinAsinB)(cosAcosB+sinAsinB)
=cosA^2*cosB^2-sinA^2*sinB^2
=cosA^2*cosB^2-(1-cosA^2)*(1-cosB^2)
=cosA^2+cosB^2-1
=1/4
所以，cosA^2+cosB^2 =5/4

已知cos(A+B)cos(A-B)=1/4
故cos(A+B)cos(A-B）=(cosAcosB-sinAsinB)(cosAcosB+sinAsinB)
=cosA^2*cosB^2-sinA^2*sinB^2
=cosA^2*cosB^2-(1-cosA^2)*(1-cosB^2)
=cosA^2+cosB^2-1
=1/4
很高兴为你解答

cos(A+B)cos(A-B) = 1/2{ cos[(A+B)+(A-B)] + cos[(A+B)-(A-B)]
= 1/2 cos2A + 1/2 cos2B
= 1/2 (2cos^2A-1) + 1/2(2cos^2B-1)
= cosA^2+cosB^2 - 1 = 1/4
cosA^2+cosB^2 = 5/4

因为cos(A+B)cos(A-B)=(1/2)(cos2A+cos2B)
=(1/2)[2(cosA)^2-1+2(cosB)^2-1]
=(cosA)^2+(cosB)^2-1=1/4
所以cosA^2+cosB^2 =1+1/4=5/4