Bài tập nâng cao · Bài 26
Bài tập nâng cao Giá trị lượng giác của góc lượng giác · Bài 26
Cho tam giác $ABC$. Chứng minh :
a. $\sin B=\sin\left( A+C \right)$.
b. $\cos\left( A+B \right)=-\cos C$.
c. $\sin\frac{A+B}{2}=\cos\frac{C}{2}$.
d. $\cos\left( B-C \right)=-\cos\left( A+2C \right)$.
e. $\cos\left( A+B-C \right)=-\cos 2C$.
f. $\cos\frac{-3A+B+C}{2}=\sin 2A$.
g. $\sin\frac{A+B+3C}{2}=\cos C$.
h. $\tan\frac{A+B-2C}{2}=\cot\frac{3C}{2}$.
a. $\sin B=\sin\left( A+C \right)$.
b. $\cos\left( A+B \right)=-\cos C$.
c. $\sin\frac{A+B}{2}=\cos\frac{C}{2}$.
d. $\cos\left( B-C \right)=-\cos\left( A+2C \right)$.
e. $\cos\left( A+B-C \right)=-\cos 2C$.
f. $\cos\frac{-3A+B+C}{2}=\sin 2A$.
g. $\sin\frac{A+B+3C}{2}=\cos C$.
h. $\tan\frac{A+B-2C}{2}=\cot\frac{3C}{2}$.
Xem lời giải
Lời giải
a. Vì $A,B,C$ là 3 góc của $\Delta ABC$ nên ta có:
$\begin{array}{l} \widehat{A}+\widehat{B}+\widehat{C}=180^{\circ} \\ \Rightarrow \widehat{B}=180^{\circ}-\left( \widehat{A}+\widehat{C} \right) \\ \Rightarrow \sin B=\sin\left[ 180^{\circ}-\left( A+C \right) \right]=\sin\left( A+C \right) \end{array}$
Vậy $\sin B=\sin\left( A+C \right)$
b. Vì $A,B,C$ là 3 góc của $\Delta ABC$ nên ta có:
$\begin{array}{l} \widehat{A}+\widehat{B}+\widehat{C}=180^{\circ} \\ \Rightarrow \widehat{A}+\widehat{B}=180^{\circ}-\widehat{C} \\ \Rightarrow \cos\left( A+B \right)=\cos\left[ 180^{\circ}-C \right]=-\cos C \end{array}$
Vậy $\cos\left( A+B \right)=-\cos C$
c. Vì $A,B,C$ là 3 góc của $\Delta ABC$ nên ta có:
$\begin{array}{l} \widehat{A}+\widehat{B}+\widehat{C}=180^{\circ} \\ \Rightarrow \widehat{A}+\widehat{B}=180^{\circ}-\widehat{C} \\ \Rightarrow \frac{\widehat{A}+\widehat{B}}{2}=\frac{180^{\circ}-\widehat{C}}{2}=90^{\circ}-\frac{\widehat{C}}{2} \\ \Rightarrow \sin\left( \frac{\widehat{A}+\widehat{B}}{2} \right)=\sin\left( 90^{\circ}-\frac{\widehat{C}}{2} \right)=\cos\frac{\widehat{C}}{2} \end{array}$
Vậy $\sin\frac{A+B}{2}=\cos\frac{C}{2}$
d. Vì $A,B,C$ là 3 góc của $\Delta ABC$ nên ta có:
$\begin{array}{l} \widehat{A}+\widehat{B}+\widehat{C}=180^{\circ} \\ \Rightarrow \widehat{B}+\widehat{C}=180^{\circ}-\widehat{A} \\ \Rightarrow \widehat{B}+\widehat{C}-2\widehat{C}=180^{\circ}-\widehat{A}-2\widehat{C} \\ \Rightarrow \widehat{B}-\widehat{C}=180^{\circ}-\left( \widehat{A}+2\widehat{C} \right) \\ \Rightarrow \cos\left( B-C \right)=\cos\left[ 180^{\circ}-\left( A+2C \right) \right]=-\cos\left( A+2C \right) \end{array}$
Vậy $\cos\left( B-C \right)=-\cos\left( A+2C \right)$
e. Vì $A,B,C$ là 3 góc của $\Delta ABC$ nên ta có:
$\begin{array}{l} \widehat{A}+\widehat{B}+\widehat{C}=180^{\circ} \\ \Rightarrow \widehat{A}+\widehat{B}=180^{\circ}-\widehat{C} \\ \Rightarrow \widehat{A}+\widehat{B}-\widehat{C}=180^{\circ}-\widehat{C}-\widehat{C} \\ \Rightarrow \widehat{A}+\widehat{B}-\widehat{C}=180^{\circ}-2\widehat{C} \\ \Rightarrow \cos\left( A+B-C \right)=\cos\left( 180^{\circ}-2C \right)=-\cos 2C \end{array}$
Vậy $\cos\left( A+B-C \right)=-\cos 2C$
f. Vì $A,B,C$ là 3 góc của $\Delta ABC$ nên ta có:
$\begin{array}{l} \widehat{A}+\widehat{B}+\widehat{C}=180^{\circ} \\ \Rightarrow \widehat{B}+\widehat{C}=180^{\circ}-\widehat{A} \\ \Rightarrow -3\widehat{A}+\widehat{B}+\widehat{C}=-3\widehat{A}+180^{\circ}-\widehat{A} \\ \Rightarrow -3\widehat{A}+\widehat{B}+\widehat{C}=180^{\circ}-4\widehat{A} \\ \Rightarrow \frac{-3\widehat{A}+\widehat{B}+\widehat{C}}{2}=\frac{180^{\circ}-4\widehat{A}}{2}=90^{\circ}-2\widehat{A} \\ \Rightarrow \cos\frac{-3A+B+C}{2}=\cos\left( 90^{\circ}-2A \right)=\sin 2A \end{array}$
Vậy $\cos\frac{-3A+B+C}{2}=\sin 2A$
g. Vì $A,B,C$ là 3 góc của $\Delta ABC$ nên ta có:
$\begin{array}{l} \widehat{A}+\widehat{B}+\widehat{C}=180^{\circ} \\ \Rightarrow \widehat{A}+\widehat{B}=180^{\circ}-\widehat{C} \\ \Rightarrow \widehat{A}+\widehat{B}+3\widehat{C}=180^{\circ}-\widehat{C}+3\widehat{C} \\ \Rightarrow \widehat{A}+\widehat{B}+3\widehat{C}=180^{\circ}+2\widehat{C} \\ \Rightarrow \frac{\widehat{A}+\widehat{B}+3\widehat{C}}{2}=\frac{180^{\circ}+2\widehat{C}}{2}=90^{\circ}+\widehat{C} \\ \Rightarrow \sin\left( \frac{A+B+3C}{2} \right)=\sin\left( 90^{\circ}+C \right)=\cos C \end{array}$
Vậy $\sin\frac{A+B+3C}{2}=\cos C$
h. Vì $A,B,C$ là 3 góc của $\Delta ABC$ nên ta có:
$\begin{array}{l} \widehat{A}+\widehat{B}+\widehat{C}=180^{\circ} \\ \Rightarrow \widehat{A}+\widehat{B}=180^{\circ}-\widehat{C} \\ \Rightarrow \widehat{A}+\widehat{B}-2\widehat{C}=180^{\circ}-\widehat{C}-2\widehat{C} \\ \Rightarrow \widehat{A}+\widehat{B}-2\widehat{C}=180^{\circ}-3\widehat{C} \\ \Rightarrow \frac{\widehat{A}+\widehat{B}-2\widehat{C}}{2}=\frac{180^{\circ}-3\widehat{C}}{2}=90^{\circ}-\frac{3\widehat{C}}{2} \\ \Rightarrow \tan\left( \frac{A+B-2C}{2} \right)=\tan\left( 90^{\circ}-\frac{3C}{2} \right)=\cot\frac{3C}{2} \end{array}$
Vậy $\tan\frac{A+B-2C}{2}=\cot\frac{3C}{2}$
$\begin{array}{l} \widehat{A}+\widehat{B}+\widehat{C}=180^{\circ} \\ \Rightarrow \widehat{B}=180^{\circ}-\left( \widehat{A}+\widehat{C} \right) \\ \Rightarrow \sin B=\sin\left[ 180^{\circ}-\left( A+C \right) \right]=\sin\left( A+C \right) \end{array}$
Vậy $\sin B=\sin\left( A+C \right)$
b. Vì $A,B,C$ là 3 góc của $\Delta ABC$ nên ta có:
$\begin{array}{l} \widehat{A}+\widehat{B}+\widehat{C}=180^{\circ} \\ \Rightarrow \widehat{A}+\widehat{B}=180^{\circ}-\widehat{C} \\ \Rightarrow \cos\left( A+B \right)=\cos\left[ 180^{\circ}-C \right]=-\cos C \end{array}$
Vậy $\cos\left( A+B \right)=-\cos C$
c. Vì $A,B,C$ là 3 góc của $\Delta ABC$ nên ta có:
$\begin{array}{l} \widehat{A}+\widehat{B}+\widehat{C}=180^{\circ} \\ \Rightarrow \widehat{A}+\widehat{B}=180^{\circ}-\widehat{C} \\ \Rightarrow \frac{\widehat{A}+\widehat{B}}{2}=\frac{180^{\circ}-\widehat{C}}{2}=90^{\circ}-\frac{\widehat{C}}{2} \\ \Rightarrow \sin\left( \frac{\widehat{A}+\widehat{B}}{2} \right)=\sin\left( 90^{\circ}-\frac{\widehat{C}}{2} \right)=\cos\frac{\widehat{C}}{2} \end{array}$
Vậy $\sin\frac{A+B}{2}=\cos\frac{C}{2}$
d. Vì $A,B,C$ là 3 góc của $\Delta ABC$ nên ta có:
$\begin{array}{l} \widehat{A}+\widehat{B}+\widehat{C}=180^{\circ} \\ \Rightarrow \widehat{B}+\widehat{C}=180^{\circ}-\widehat{A} \\ \Rightarrow \widehat{B}+\widehat{C}-2\widehat{C}=180^{\circ}-\widehat{A}-2\widehat{C} \\ \Rightarrow \widehat{B}-\widehat{C}=180^{\circ}-\left( \widehat{A}+2\widehat{C} \right) \\ \Rightarrow \cos\left( B-C \right)=\cos\left[ 180^{\circ}-\left( A+2C \right) \right]=-\cos\left( A+2C \right) \end{array}$
Vậy $\cos\left( B-C \right)=-\cos\left( A+2C \right)$
e. Vì $A,B,C$ là 3 góc của $\Delta ABC$ nên ta có:
$\begin{array}{l} \widehat{A}+\widehat{B}+\widehat{C}=180^{\circ} \\ \Rightarrow \widehat{A}+\widehat{B}=180^{\circ}-\widehat{C} \\ \Rightarrow \widehat{A}+\widehat{B}-\widehat{C}=180^{\circ}-\widehat{C}-\widehat{C} \\ \Rightarrow \widehat{A}+\widehat{B}-\widehat{C}=180^{\circ}-2\widehat{C} \\ \Rightarrow \cos\left( A+B-C \right)=\cos\left( 180^{\circ}-2C \right)=-\cos 2C \end{array}$
Vậy $\cos\left( A+B-C \right)=-\cos 2C$
f. Vì $A,B,C$ là 3 góc của $\Delta ABC$ nên ta có:
$\begin{array}{l} \widehat{A}+\widehat{B}+\widehat{C}=180^{\circ} \\ \Rightarrow \widehat{B}+\widehat{C}=180^{\circ}-\widehat{A} \\ \Rightarrow -3\widehat{A}+\widehat{B}+\widehat{C}=-3\widehat{A}+180^{\circ}-\widehat{A} \\ \Rightarrow -3\widehat{A}+\widehat{B}+\widehat{C}=180^{\circ}-4\widehat{A} \\ \Rightarrow \frac{-3\widehat{A}+\widehat{B}+\widehat{C}}{2}=\frac{180^{\circ}-4\widehat{A}}{2}=90^{\circ}-2\widehat{A} \\ \Rightarrow \cos\frac{-3A+B+C}{2}=\cos\left( 90^{\circ}-2A \right)=\sin 2A \end{array}$
Vậy $\cos\frac{-3A+B+C}{2}=\sin 2A$
g. Vì $A,B,C$ là 3 góc của $\Delta ABC$ nên ta có:
$\begin{array}{l} \widehat{A}+\widehat{B}+\widehat{C}=180^{\circ} \\ \Rightarrow \widehat{A}+\widehat{B}=180^{\circ}-\widehat{C} \\ \Rightarrow \widehat{A}+\widehat{B}+3\widehat{C}=180^{\circ}-\widehat{C}+3\widehat{C} \\ \Rightarrow \widehat{A}+\widehat{B}+3\widehat{C}=180^{\circ}+2\widehat{C} \\ \Rightarrow \frac{\widehat{A}+\widehat{B}+3\widehat{C}}{2}=\frac{180^{\circ}+2\widehat{C}}{2}=90^{\circ}+\widehat{C} \\ \Rightarrow \sin\left( \frac{A+B+3C}{2} \right)=\sin\left( 90^{\circ}+C \right)=\cos C \end{array}$
Vậy $\sin\frac{A+B+3C}{2}=\cos C$
h. Vì $A,B,C$ là 3 góc của $\Delta ABC$ nên ta có:
$\begin{array}{l} \widehat{A}+\widehat{B}+\widehat{C}=180^{\circ} \\ \Rightarrow \widehat{A}+\widehat{B}=180^{\circ}-\widehat{C} \\ \Rightarrow \widehat{A}+\widehat{B}-2\widehat{C}=180^{\circ}-\widehat{C}-2\widehat{C} \\ \Rightarrow \widehat{A}+\widehat{B}-2\widehat{C}=180^{\circ}-3\widehat{C} \\ \Rightarrow \frac{\widehat{A}+\widehat{B}-2\widehat{C}}{2}=\frac{180^{\circ}-3\widehat{C}}{2}=90^{\circ}-\frac{3\widehat{C}}{2} \\ \Rightarrow \tan\left( \frac{A+B-2C}{2} \right)=\tan\left( 90^{\circ}-\frac{3C}{2} \right)=\cot\frac{3C}{2} \end{array}$
Vậy $\tan\frac{A+B-2C}{2}=\cot\frac{3C}{2}$