知cost/(1-sint)=1/[(pai/1)-1]如何得出tan(t/2)=(pai-2)/2

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知cost/(1-sint)=1/[(pai/1)-1]如何得出tan(t/2)=(pai-2)/2知cost/(1-sint)=1/[(pai/1)-1]如何得出tan(t/2)=(pai-2)/2

知cost/(1-sint)=1/[(pai/1)-1]如何得出tan(t/2)=(pai-2)/2
知cost/(1-sint)=1/[(pai/1)-1]如何得出tan(t/2)=(pai-2)/2

知cost/(1-sint)=1/[(pai/1)-1]如何得出tan(t/2)=(pai-2)/2
cost/(1-sint)
=(cos^2(t/2)-sin^2(t/2))/(sin^2(t/2)+cos^2(t/2)-2sin(t/2)cos(t/2))
=(1-tan^2(t/2))/(tan^2(t/2)+1-2tan(t/2))
=1/(1/π-1)
(1-tan^2(t/2))(1/π-1)=tan^2(t/2)+1-2tan(t/2)
令tan(t/2)=x
(1-x^2)(1/π-1)=x^2-2x+1
1/π-1-x^2/π+x^2=x^2-2x+1
1/π-x^2/π=-2x+2
x^2-2πx+2π-1=0
(x-π)^2=(π-1)^2
x-π=π-1 (1-π舍去)
x=2π-1
所以tan(t/2)=2π-1