Cho tam giác ABC vuông tại A. $CMR: tan\frac{B}{2}=\frac{AC}{AB+BC}$. Áp dụng tính tan$15^{0}$
$CMR: tan\frac{B}{2}=\frac{AC}{AB+BC}$
Started By nguyennamphu1810, 03-08-2015 - 10:19
#1
Posted 03-08-2015 - 10:19
#2
Posted 03-08-2015 - 11:05
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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