求sin∧2a·sin∧2β+cos∧2a·cos∧2β-1/2cos2a·cos2β..
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求sin∧2a·sin∧2β+cos∧2a·cos∧2β-1/2cos2a·cos2β..
(sina)^2*(sinb)^2+(cosa)^2*(cosb)^2-(1/2)(cos2acos2b)
=(sinasinb)^2+(cosacosb)^2-(1/2)[cos2acos2b)
=(1/4)(cos(a-b)-cos(a+b))^2+(1/4)cos(a-b)+cos(a+b))^2-(1/4)cos(2a+2b)-(1/4)cos(2a-2b)
=(1/2)cos(a+b)^2+(1/2)cos(a-b)^2-(1/4)(2cos(a+b)^2-1)-(1/4)(2cos(a-b)^2-1)
=1/4+1/4
=1/2
=(sinasinb)^2+(cosacosb)^2-(1/2)[cos2acos2b)
=(1/4)(cos(a-b)-cos(a+b))^2+(1/4)cos(a-b)+cos(a+b))^2-(1/4)cos(2a+2b)-(1/4)cos(2a-2b)
=(1/2)cos(a+b)^2+(1/2)cos(a-b)^2-(1/4)(2cos(a+b)^2-1)-(1/4)(2cos(a-b)^2-1)
=1/4+1/4
=1/2
求sin∧2a·sin∧2β+cos∧2a·cos∧2β-1/2cos2a·cos2β..
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