Cho tam giác ABC có BC = a, AC = b, AB = c. Chứng minh rằng \(\sin \frac{A}{2} \le \frac{a}{{b + c}}\).

A. 1.

B. 1 + 2sinxcosx.

C. 1 – 2sinxcosx.

D. 0.

A. 3sin²xcos²x.

B. sin²x.

C. 1 – 3sin²xcos²x.

D. 2 + sin²x.

A. \(\frac{{1 + \cot x}}{{1 - \cot x}} = \frac{{\tan x + 1}}{{\tan x - 1}}\).

B. \(\frac{{1 + \cot x}}{{1 - \cot x}} = \frac{{\tan x}}{{\tan x - 1}}\).

C. \(\frac{{1 + \cot x}}{{1 - \cot x}} = \frac{{\tan x + 1}}{{\tan x}}\).

D. \(\frac{{1 + \cot x}}{{1 - \cot x}} = \frac{{{{\tan }^2}x + 1}}{{\tan x - 1}}\).

A. sinA = sin(B + C).

B. tanA = tan(B + C).

C. cos\(\frac{A}{2}\) = sin\(\frac{{B + C}}{2}\) .

D. tanA = −tan(B + C).

A. \({\cos ^2}\frac{\alpha }{2} + {\sin ^2}\frac{\alpha }{2} = \frac{1}{2}\).

B. \({\cos ^2}\frac{\alpha }{3} + {\sin ^2}\frac{\alpha }{3} = \frac{1}{3}\).

C. \({\cos ^2}\frac{\alpha }{4} + {\sin ^2}\frac{\alpha }{4} = \frac{1}{4}\).

D. \(5\left( {{{\cos }^2}\frac{\alpha }{5} + {{\sin }^2}\frac{\alpha }{5}} \right) = 5\).