Question

Cho ΔABC vuông tại A.

a) CM: \(\tan\frac{B}{2}=\frac{AC}{AB+BC}\)

b) Tính \(\tan15^o\)

c) Tính \(\tan22^o30'\)

Answer 1

a: \(tan\left(\dfrac{B}{2}\right)=\dfrac{2\cdot\dfrac{sinB}{2}\cdot cos\left(\dfrac{B}{2}\right)}{2\cdot cos^2\left(\dfrac{B}{2}\right)}\)

\(=\dfrac{sinB}{1+cosB}=\dfrac{AC}{AB+BC}\)

b: \(tan15^0=2-\sqrt{3}\)

c:góc B=45 độ

=>sin B=căn 2/2; cosB=căn 2/2

\(tan\left(\dfrac{45}{2}\right)=tan\left(\dfrac{B}{2}\right)=\dfrac{sinB}{1+cosB}=\dfrac{\sqrt{2}}{2}:\dfrac{\sqrt{2}+2}{2}=\dfrac{\sqrt{2}}{2+\sqrt{2}}\)