tan(2a-β)=tan[a+(a-β)]=[tana+tan(a-β)]/[1-tanatan(a-β)]
=(1/2-2/5)/(1+1/2*2/5)
=(1/10)/(6/5)
=1/12
tan(β-2a)=-tan(2a-β)=-1/12
已知tana=1/2,tan(a-p)=-2/5,那么tan(p-2a)的值为
已知tana=1/2,tan(a-p)=-2/5,那么tan(p-2a)的值为
数学人气:633 ℃时间:2020-02-03 18:32:44
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