cos(a+b)cos(a-b)=1/5,则(cosa)^2-(sinb)^2=?
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cos(a+b)cos(a-b)=1/5,则(cosa)^2-(sinb)^2=?
cos(a+b)cos(a-b)=1/5,则(cosa)^2-(sinb)^2=?
cos(a+b)cos(a-b)=1/5,则(cosa)^2-(sinb)^2=?
1/5.
cos(a+b)cos(a-b)=(cosacosb-sinasinb)(cosacosb+sinasinb)
=(cosacosb)^2-(sinasinb)^2
=cosa^2[1-(sinab)^2]-sinb^2[1-(cosa)^2]
=(cosa)^2-(sinb)^2
=1/5
(cosa)^2-(sinb)^2=cos2a
2a=(a+b)+(a-b)
cos2a=cos[(a+b)+(a-b)]=cos(a+b)cos(a-b)-sin(a+b)sin(a-b)
cos(a+b)cos(a-b)=1/5 sin(a+b)sin(a-b)=1-1\25开方=2倍根号下6\5
所以(cosa)^2-(sinb)^2=cos2a=co...
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(cosa)^2-(sinb)^2=cos2a
2a=(a+b)+(a-b)
cos2a=cos[(a+b)+(a-b)]=cos(a+b)cos(a-b)-sin(a+b)sin(a-b)
cos(a+b)cos(a-b)=1/5 sin(a+b)sin(a-b)=1-1\25开方=2倍根号下6\5
所以(cosa)^2-(sinb)^2=cos2a=cos(a+b)cos(a-b)-sin(a+b)sin(a-b)=【1-2倍根号下6】|5
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