Question

Cho A,B,C là ba góc của một tam giác . Chứng minh rằng :

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

b/ \(cos\left(A+B\right)=-cosC\)

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

d/ sinA=sin(B+C)

e/ sin(A+B)=sinC

f/ cosA=-cos(B+C)

Answer 1

\(A+B+C=180^0\Rightarrow\frac{A+B}{2}+\frac{C}{2}=90^0\)

\(\Rightarrow sin\left(\frac{A+B}{2}\right)=cos\left(90^0-\frac{A+B}{2}\right)=cos\frac{C}{2}\)

\(cos\left(A+B\right)=-cos\left(180^0-\left(A+B\right)\right)=-cosC\)

\(cos\left(\frac{A+B}{2}\right)=sin\left(90-\frac{A+B}{2}\right)=sin\frac{C}{2}\)

\(sinA=sin\left(180^0-A\right)=sin\left(B+C\right)\)

\(sin\left(A+B\right)=sin\left(180^0-\left(A+B\right)\right)=sinC\)

\(cosA=-cos\left(180^0-A\right)=-cos\left(B+C\right)\)