\(a,sin^2A=sinB.sinC\)

\(\Leftrightarrow\frac{a^2}{4R^2}=\frac{b}{2R}.\frac{c}{2R}\)

\(\Leftrightarrow\frac{a^2}{4R}=\frac{bc}{4R^2}\Leftrightarrow a^2=bc\)

b, Áp dụng định lý cos:

\(CosA=\frac{b^2+c^2-a^2}{2bc}=\frac{b^2+c^2-bc}{2bc}\ge\frac{2bc-bc}{2bc}=-\frac{1}{2}\)

Trong một tam giác thì tổng các góc là 180^{0 :}

+ + = 180^{0 => = -1800 - ( + )}

và ( + ) là 2 góc bù nhau, do đó:

a) sinA = sin[180⁰ - ( + )] = sin (B + C)

b) cosA = cos[180⁰ - ( + )] = -cos (B + C)

a) Sin (B+C) = Sin (180-A) = Sin A
b) Cos (A+B) = Cos ( 180-A) = Cos A
c) Sin (\(\dfrac{B+C}{2}\)) = Sin \(\left(\dfrac{180-A}{2}\right)\)= Sin \(\left(90^0-\dfrac{A}{2}\right)\)= Cos \(\dfrac{A}{2}\)

d) Tan \(\left(\dfrac{A+C}{2}\right)\)= Tan\(\left(\dfrac{180-B}{2}\right)\)=Tan\(\left(90^0-\dfrac{B}{2}\right)\)= Cot \(\dfrac{B}{2}\)

a) ta có : A+B+C=180=\(\pi\)

=>B+C= \(\pi\) - A

=> sin (B+C)=Sin(\(\pi\)-A)=SinA

b) tương tự:

cos( A+B)= Cos (\(\pi\)-C)=-cosC

c) ta có A+B+C =\(\pi\)=>\(\frac{A}{2}\)+\(\frac{B}{2}\)+\(\frac{C}{2}\)=\(\frac{\pi}{2}\)

=> sin (\(\frac{B+C}{2}\))=sin(\(\frac{\pi}{2}\)-\(\frac{A}{2}\))=cos(\(\frac{A}{2}\))

d) tương tự:

tan \(\frac{A+C}{2}\)=tan(\(\frac{\pi}{2}\)-\(\frac{B}{2}\))= cot\(\frac{B}{2}\)

===> đpcm