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已知sin(a+b)*sin(a-b)=1/3,求:(1/4)(sin2a)^2+(sinb)^2+(cosa)^4=?
人气:301 ℃ 时间:2020-06-16 17:08:56
解答
1/4sin2a.sin2a+sinb.sinb+cosa.cosa.cosa.cosa
=sina^2*cosa^2+sinb^2+cosa^2*cosa^2
=cosa^2(sina^2+cosa^2)+sinb^2
=cosa^2+sinb^2
sin(a+b)sin(a-b)=1/3
(sinacosb+cosasinb)(sinacosb-cosasinb)=1/3
sina^2cosb^2-cosa^2sinb^2=1/3
(1-cosa^2)(1-sinb^2)-cosa^2sinb^2=1/3
[1+cosa^2sinb^2-(cosa^2+sinb^2)]-cosa^2sinb^2=1/3
1-(cosa^2+sinb^2)=1/3
cosa^2+sinb^2=2/3
所以1/4sin2a.sin2a+sinb.sinb+cosa.cosa.cosa.cosa=2/3
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