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$CMR: tan\frac{B}{2}=\frac{AC}{AB+BC}$


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#1
nguyennamphu1810

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Cho tam giác ABC vuông tại A. $CMR: tan\frac{B}{2}=\frac{AC}{AB+BC}$. Áp dụng tính tan$15^{0}$



#2
Minhnguyenthe333

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Cho tam giác ABC vuông tại A. $CMR: tan\frac{B}{2}=\frac{AC}{AB+BC}$. Áp dụng tính tan$15^{0}$

Đặt $k=tan\frac{B}{2}$.Ta có:

$\frac{1}{tanB}=\frac{1-k^2}{2k}=\frac{1}{2k}-\frac{k}{2}\Leftrightarrow \frac{1}{tanB}+\frac{1}{2k}+\frac{k}{2}=\frac{1}{k}$

Lại có:$\frac{1}{sinB}=\frac{1+k^2}{2k}=\frac{1}{2k}+\frac{k}{2}$

$\Rightarrow  \frac{1}{tanB}+\frac{1}{sinB}=\frac{1}{k}\Leftrightarrow \frac{AB}{AC}+\frac{BC}{AC}=\frac{1}{k}$

$\Rightarrow tan\frac{B}{2}=\frac{AC}{AB+BC}$

Ta có:$\frac{1}{tan 15^0}=\frac{1}{tan30}+\frac{1}{sin30}=2+\sqrt{3}\Rightarrow tan15=\frac{1}{2+\sqrt{3}}$


Edited by Minhnguyenthe333, 03-08-2015 - 11:06.





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