tan(-14π/15)=-tan(1-1π/15)=tan1π/15对吗?
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tan(-14π/15)=-tan(1-1π/15)=tan1π/15对吗?tan(-14π/15)=-tan(1-1π/15)=tan1π/15对吗?tan(-14π/15)=-tan(1-1π/1
tan(-14π/15)=-tan(1-1π/15)=tan1π/15对吗?
tan(-14π/15)=-tan(1-1π/15)=tan1π/15对吗?
tan(-14π/15)=-tan(1-1π/15)=tan1π/15对吗?
tan(-14π/15)=-tan(π-1π/15)=tan1π/15看明白了吗?
不对
tan(-14π/15)=-tan(π/15)=tan(-π/15)=a 因为tan(-π/15)=sin(-π/15)/cos(-π/15) 设sin(-π/15)=x 所以x/根号(1-x^2)=a x
三角函数算值题1-tan^2(π/8) /tan(π/8)=tan(15)/1-tan^2(15)=
tan(-14π/15)=-tan(π/15)=tan(-π/15)对吗?
tan(-14π/15)=-tan(1-1π/15)=tan1π/15对吗?
tan（ x/2+π/4）+tan（x/2-π/4 ）=2tanxtan(x/2+π/4)+tan(x/2-π/4)=[tan(x/2)+tan(π/4)]/[1-tan(x/2)tan(π/4)]+[tan(x/2)-tan(π/4)]/[1+tan(x/2)tan(π/4)]=[tan(x/2)+1]/[1-tan(x/2)]+[tan(x/2)-1]/[1+tan(x/2)]=[(tan(x/2)+1)^2-(tan(x/2)-1)^2]/[1-(tan(x
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tan(π/2+1)=?tan(1-π/2)=?
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tan(π/12)+1/tan(π/12)=啥?
tanπ/12-1/tanπ/12=
这道题为什么第二步的+到了第三步变成了-1.tan(x/2+π/4)+tan(x/2-π/4)=[tan(x/2)+tan(π/4)]/[1-tan(x/2)tan(π/4)]+[tan(x/2)-tan(π/4)]/[1+tan(x/2)tan(π/4)]=[tan(x/2)+1]/[1-tan(x/2)]+[tan(x/2)-1]/[1+tan(x/2)]=[(tan(x/2)+1)^2-(tan(x/2)-
tan(π/12)-(1/tan(π/12))
tan=3,π
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