Cho tam giác ABC, chứng minh: sinA + sinB + sinC = 4cosA/2.cosB/2.cosC/2

Đề bài: Cho tam giác ABC, chứng minh:

$sin\widehat{A}+\sin \widehat{B}+\sin \widehat{C}=4.\cos \frac{\widehat{A}}{2}.\cos \frac{\widehat{B}}{2}.\cos \frac{\widehat{C}}{2}$.

Trả lời

Hướng dẫn giải:

Ta có: $\widehat{A}+\widehat{B}+\widehat{C}=180°$

$\Rightarrow \sin \frac{(\widehat{A}+\widehat{B})}{2}=\sin \left(\frac{180°}{2}-\frac{\widehat{C}}{2}\right)=\cos \frac{\widehat{C}}{2}$

Tương tự ta có: $\sin \frac{\widehat{C}}{2}=\cos \frac{\widehat{A}+\widehat{B}}{2}$

$$\begin{gathered} \Rightarrow \sin \widehat{A}+\sin \widehat{B}+\sin \widehat{C}=2\cos \frac{\widehat{C}}{2}.\cos \frac{\widehat{A}-\widehat{B}}{2}+2\cos \frac{\widehat{A}+\widehat{B}}{2}.\cos \frac{\widehat{C}}{2} \\ =2\cos \frac{\widehat{C}}{2}\left(\cos \frac{\widehat{A}-\widehat{B}}{2}+\cos \frac{\widehat{A}+\widehat{B}}{2}\right) \end{gathered}$$

$=4.\cos \frac{\widehat{A}}{2}.\cos \frac{\widehat{B}}{2}.\cos \frac{\widehat{C}}{2}$ (đpcm).