Bài tập nâng cao · Bài 18
Dạng 5. Bài toán tam giác Qui ước: Cho tam giác $ABC$ gọi $a,b,c$ là ba cạnh đối diện của ba góc $A,B,C$; $h_a,h_b,h_c$ là ba đường cao; $m_a,m_b,m_c$ là ba đường trung tuyến; $l_A,l_B,l_C$ là ba đường phân giác; $r$ là bán kính đường trong nội tiếp; $R$ là bán kính đường trong ngoại tiếp và $p=\frac{a+b+c}{2}$ là nữa chu vi. Điều kiện $A,B,C$ là ba góc của một tam giác là $\left\{ \begin{array}{l} A,B,C \\ A+B+C=\pi \end{array} \right.$ nên suy ra $A+B=\pi -C$, $\frac{A+B}{2}=\frac{\pi}{2}-\frac{C}{2}$… Định lý hàm số côsin $a^{2}=b^{2}+c^{2}-2bc\cos A$, $b^{2}=a^{2}+c^{2}-2ac\cos B$, $c^{2}=a^{2}+b^{2}-2ab\cos C$ Suy ra $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc}$, $\cos B=\frac{a^{2}+c^{2}-b^{2}}{2ac}$, $\cos C=\frac{a^{2}+b^{2}-c^{2}}{2ab}$ Định lý hàm số sin: $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$ suy ra $a=2R\sin A,\,b=2R\sin B,\,c=2R\sin C$ Công thưc tính diện tích $S=\frac{1}{2}ah_a=\frac{1}{2}ab\sin C=\frac{abc}{4R}=pr=\sqrt{p\left( p-a \right)\left( p-b \right)\left( p-c \right)}$. Công thức phân giác $l_A=\frac{2bc\cos\frac{A}{2}}{b+c},\ldots$ Công thức trung tuyến $m_a^{2}=\frac{b^{2}+c^{2}}{2}-\frac{a^{2}}{4},\ldots$
Bài tập nâng cao Công thức lượng giác · Bài 18
Cho tam giác $ABC$. Chứng minh rằng:
a) $\sin C=\sin A.\cos B+\sin B.\cos A$.
b) $\frac{\sin C}{\cos A.\cos B}=\tan A+\tan B\left( A,B\neq 90^{\circ} \right)$.
c) $\tan A+\tan B+\tan C=\tan A.\tan B.\tan C\left( A,B,C\neq 90^{\circ} \right)$.
d) $\cot A.\cot B+\cot B.\cot C+\cot C.\cot A=1$.
e) $\tan\frac{A}{2}.\tan\frac{B}{2}+\tan\frac{B}{2}.\tan\frac{C}{2}+\tan\frac{C}{2}.\tan\frac{A}{2}=1$.
f) $\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}=\cot\frac{A}{2}.\cot\frac{B}{2}.\cot\frac{C}{2}$.
g) $\cot B+\frac{\cos C}{\sin B.\cos A}=\cot C+\frac{\cos B}{\sin C.\cos A}\left( A\neq 90^{\circ} \right)$.
h) $\cos\frac{A}{2}.\cos\frac{B}{2}.\cos\frac{C}{2}=\sin\frac{A}{2}.\sin\frac{B}{2}.\cos\frac{C}{2}+\sin\frac{A}{2}.\cos\frac{B}{2}.\sin\frac{C}{2}+\cos\frac{A}{2}.\sin\frac{B}{2}.\sin\frac{C}{2}$.
i) $\sin^{2}\frac{A}{2}+\sin^{2}\frac{B}{2}+\sin^{2}\frac{C}{2}=1-2\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$.
a) $\sin C=\sin A.\cos B+\sin B.\cos A$.
b) $\frac{\sin C}{\cos A.\cos B}=\tan A+\tan B\left( A,B\neq 90^{\circ} \right)$.
c) $\tan A+\tan B+\tan C=\tan A.\tan B.\tan C\left( A,B,C\neq 90^{\circ} \right)$.
d) $\cot A.\cot B+\cot B.\cot C+\cot C.\cot A=1$.
e) $\tan\frac{A}{2}.\tan\frac{B}{2}+\tan\frac{B}{2}.\tan\frac{C}{2}+\tan\frac{C}{2}.\tan\frac{A}{2}=1$.
f) $\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}=\cot\frac{A}{2}.\cot\frac{B}{2}.\cot\frac{C}{2}$.
g) $\cot B+\frac{\cos C}{\sin B.\cos A}=\cot C+\frac{\cos B}{\sin C.\cos A}\left( A\neq 90^{\circ} \right)$.
h) $\cos\frac{A}{2}.\cos\frac{B}{2}.\cos\frac{C}{2}=\sin\frac{A}{2}.\sin\frac{B}{2}.\cos\frac{C}{2}+\sin\frac{A}{2}.\cos\frac{B}{2}.\sin\frac{C}{2}+\cos\frac{A}{2}.\sin\frac{B}{2}.\sin\frac{C}{2}$.
i) $\sin^{2}\frac{A}{2}+\sin^{2}\frac{B}{2}+\sin^{2}\frac{C}{2}=1-2\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$.
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Lời giải
j) $\sin C=\sin A.\cos B+\sin B.\cos A$.
Ta có $A+B+C=\pi \Leftrightarrow A+B=\pi -C$.
$\Leftrightarrow \sin\left( A+B \right)=\sin\left( \pi -C \right)$. $\Leftrightarrow$ $\sin A.\cos B+\sin B.\cos A=\sin C$ (đpcm).
k) $\frac{\sin C}{\cos A.\cos B}=\tan A+\tan B\,\left( A,\,B\neq 90^{\circ} \right)$.
Ta có $\tan A+\tan B=\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B}=\frac{\sin A\cos B+\cos A\sin B}{\cos A.\cos B}=\frac{\sin\left( A+B \right)}{\cos A.\cos B}=\frac{\sin C}{\cos A.\cos B}$.
l) $\tan A+\tan B+\tan C=\tan A.\tan B.\tan C\,\left( A,\,B,\,C\neq 90^{\circ} \right)$
Ta có $-\tan C=\tan\left( A+B \right)=\frac{\tan A+\tan B}{1-\tan A.\tan B}$.
Nên $-\tan C\left( 1-\tan A.\tan B \right)=\tan A+\tan B$.
Do đó $\tan A+\tan B+\tan C=\tan A.\tan B.\tan C$.
m) $\cot A.\cot B+\cot B.\cot C+\cot C.\cot A=1$.
$\cot A.\cot B+\cot B.\cot C+\cot C.\cot A=\cot B\left( \cot A+\cot C \right)+\cot C.\cot A$.
$=\frac{\cos B}{\sin B}\left( \frac{\sin\left( A+C \right)}{\sin A.\sin C} \right)+\cot C.\cot A$.
$=\frac{\cos B}{\sin A.\sin C}+\frac{\cos C.\cos A}{\sin A.\sin C}=\frac{\cos B+\frac{1}{2}\left( \cos\left( A+C \right)+\cos\left( A-C \right) \right)}{\sin A.\sin C}$.
$=\frac{\frac{1}{2}\left( \cos\left( A-C \right)-\cos\left( A+C \right) \right)}{\sin A.\sin C}=\frac{\sin A.\sin C}{\sin A.\sin C}=1$.
n) $\tan\frac{A}{2}.\tan\frac{B}{2}+\tan\frac{B}{2}.\tan\frac{C}{2}+\tan\frac{C}{2}.\tan\frac{A}{2}=1$.
$\tan\frac{A}{2}.\tan\frac{B}{2}+\tan\frac{B}{2}.\tan\frac{C}{2}+\tan\frac{C}{2}.\tan\frac{A}{2}=\tan\frac{B}{2}\left( \tan\frac{A}{2}+\tan\frac{C}{2} \right)+\tan\frac{C}{2}.\tan\frac{A}{2}$.
$=\frac{\sin\frac{B}{2}}{\cos\frac{B}{2}}\left( \frac{\sin\left( \frac{A}{2}+\frac{C}{2} \right)}{\cos\frac{A}{2}\cos\frac{C}{2}} \right)+\tan\frac{C}{2}.\tan\frac{A}{2}$.
$=\frac{\sin\frac{B}{2}}{\cos\frac{A}{2}\cos\frac{C}{2}}+\frac{\sin\frac{C}{2}.\sin\frac{A}{2}}{\cos\frac{A}{2}\cos\frac{C}{2}}$.
$=\frac{\sin\frac{B}{2}+\sin\frac{C}{2}.\sin\frac{A}{2}}{\cos\frac{A}{2}\cos\frac{C}{2}}$.
$=\frac{\sin\frac{B}{2}+\frac{1}{2}\left( \cos\left( \frac{C}{2}-\frac{A}{2} \right)-\cos\left( \frac{C}{2}+\frac{A}{2} \right) \right)}{\cos\frac{A}{2}\cos\frac{C}{2}}$. $=\frac{\frac{1}{2}\left( \cos\left( \frac{C}{2}-\frac{A}{2} \right)+\cos\left( \frac{C}{2}+\frac{A}{2} \right) \right)}{\cos\frac{A}{2}\cos\frac{C}{2}}=\frac{\cos\frac{A}{2}\cos\frac{C}{2}}{\cos\frac{A}{2}\cos\frac{C}{2}}=1$.
o) $\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}=\cot\frac{A}{2}.\cot\frac{B}{2}.\cot\frac{C}{2}$.
$\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}=\cot\frac{A}{2}.\cot\frac{B}{2}.\cot\frac{C}{2}$.
$\Leftrightarrow \cot\frac{A}{2}+\cot\frac{B}{2}=\cot\frac{A}{2}.\cot\frac{B}{2}.\cot\frac{C}{2}-\cot\frac{C}{2}$.
$\Leftrightarrow \cot\frac{A}{2}+\cot\frac{B}{2}=\left( \cot\frac{A}{2}.\cot\frac{B}{2}-1 \right)\cot\frac{C}{2}$.
$\Leftrightarrow \cot\frac{A}{2}+\cot\frac{B}{2}=\left( \frac{\cos\frac{A}{2}.\cos\frac{B}{2}}{\sin\frac{A}{2}.\sin\frac{B}{2}}-1 \right)\frac{\cos\frac{C}{2}}{\sin\frac{C}{2}}$.
$\Leftrightarrow \cot\frac{A}{2}+\cot\frac{B}{2}=\left( \frac{\cos\left( \frac{A}{2}+\frac{B}{2} \right)}{\sin\frac{A}{2}.\sin\frac{B}{2}} \right)\frac{\cos\frac{C}{2}}{\sin\frac{C}{2}}$.
$\Leftrightarrow \cot\frac{A}{2}+\cot\frac{B}{2}=\frac{\cos\frac{C}{2}}{\sin\frac{A}{2}.\sin\frac{B}{2}}$.
$\Leftrightarrow \frac{\cos\frac{A}{2}}{\sin\frac{A}{2}}+\frac{\cos\frac{B}{2}}{\sin\frac{B}{2}}=\frac{\cos\frac{C}{2}}{\sin\frac{A}{2}.\sin\frac{B}{2}}\Leftrightarrow \frac{\cos\frac{A}{2}\sin\frac{B}{2}+\cos\frac{B}{2}\sin\frac{A}{2}}{\sin\frac{A}{2}\sin\frac{B}{2}}=\frac{\cos\frac{C}{2}}{\sin\frac{A}{2}.\sin\frac{B}{2}}$.
$\Leftrightarrow \frac{\sin\left( \frac{A}{2}+\frac{B}{2} \right)}{\sin\frac{A}{2}\sin\frac{B}{2}}=\frac{\cos\frac{C}{2}}{\sin\frac{A}{2}.\sin\frac{B}{2}}$. Luôn đúng
Vậy $\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}=\cot\frac{A}{2}.\cot\frac{B}{2}.\cot\frac{C}{2}$.
p) $\cot B+\frac{\cos C}{\sin B.\cos A}=\cot C+\frac{\cos B}{\sin C.\cos A}\,\left( A\neq 90^{\circ} \right)$
$\cot B+\frac{\cos C}{\sin B.\cos A}=\cot C+\frac{\cos B}{\sin C.\cos A}\Leftrightarrow \cot B-\cot C=\frac{\cos B}{\sin C.\cos A}-\frac{\cos C}{\sin B.\cos A}$.
$\Leftrightarrow \frac{\cos B\sin C-\cos C\sin B}{\sin B\sin C}=\frac{1}{\cos A}\left( \frac{\frac{1}{2}\left( \sin 2B-\sin 2C \right)}{\sin B\sin C} \right)$.
$\Leftrightarrow \frac{\sin\left( C-B \right)}{\sin B\sin C}=\frac{1}{\cos A}\left( \frac{\frac{1}{2}\left( \sin 2B-\sin 2C \right)}{\sin B\sin C} \right)\Leftrightarrow \frac{\sin\left( C-B \right)}{\sin B\sin C}=\frac{1}{\cos A}\frac{\cos\left( B+C \right)\sin\left( B-C \right)}{\sin B\sin C}\Leftrightarrow \frac{\sin\left( C-B \right)}{\sin B\sin C}=\frac{1}{\cos A}\frac{-\cos A\sin\left( B-C \right)}{\sin B\sin C}\Leftrightarrow \frac{\sin\left( C-B \right)}{\sin B\sin C}=\frac{\sin\left( C-B \right)}{\sin B\sin C}$
Vậy $\cot B+\frac{\cos C}{\sin B.\cos A}=\cot C+\frac{\cos B}{\sin C.\cos A}\,\left( A\neq 90^{\circ} \right)$.
q) $\cos\frac{A}{2}.\cos\frac{B}{2}.\cos\frac{C}{2}=\sin\frac{A}{2}.\sin\frac{B}{2}.\cos\frac{C}{2}+\sin\frac{A}{2}.\cos\frac{B}{2}.\sin\frac{C}{2}+\cos\frac{A}{2}.\sin\frac{B}{2}.\sin\frac{C}{2}$.
Đặt $\sin\frac{A}{2}.\sin\frac{B}{2}.\cos\frac{C}{2}+\sin\frac{A}{2}.\cos\frac{B}{2}.\sin\frac{C}{2}+\cos\frac{A}{2}.\sin\frac{B}{2}.\sin\frac{C}{2}=T$.
$T=\sin\frac{A}{2}.\sin\left( \frac{B+C}{2} \right)+\cos\frac{A}{2}.\sin\frac{B}{2}.\sin\frac{C}{2}$.
$T=\cos\frac{A}{2}.\left( \sin\frac{A}{2}+\sin\frac{B}{2}.\sin\frac{C}{2} \right)$.
$T=\cos\frac{A}{2}.\left( \cos\frac{B+C}{2}+\sin\frac{B}{2}.\sin\frac{C}{2} \right)$.
$T=\cos\frac{A}{2}.\left( \cos\frac{B}{2}\cos\frac{C}{2}-\sin\frac{B}{2}.\sin\frac{C}{2}+\sin\frac{B}{2}.\sin\frac{C}{2} \right)=\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$.
r) $\sin^{2}\frac{A}{2}+\sin^{2}\frac{B}{2}+\sin^{2}\frac{C}{2}=1-2\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$
Ta có $\sin^{2}\frac{A}{2}+\sin^{2}\frac{B}{2}+\sin^{2}\frac{C}{2}=\frac{1-\cos A}{2}+\frac{1-\cos B}{2}+\sin^{2}\frac{C}{2}$.
$=1-\frac{1}{2}\left( \cos A+\cos B \right)+\sin^{2}\frac{C}{2}$ $=1-\frac{1}{2}\left( 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} \right)+\sin^{2}\frac{C}{2}=1-\left( \sin\frac{C}{2}\cos\frac{A-B}{2} \right)+\sin^{2}\frac{C}{2}$
$=1+\sin\frac{C}{2}\left( \cos\frac{A+B}{2}-\cos\frac{A-B}{2} \right)$
$=1-2\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$.
Ta có $A+B+C=\pi \Leftrightarrow A+B=\pi -C$.
$\Leftrightarrow \sin\left( A+B \right)=\sin\left( \pi -C \right)$. $\Leftrightarrow$ $\sin A.\cos B+\sin B.\cos A=\sin C$ (đpcm).
k) $\frac{\sin C}{\cos A.\cos B}=\tan A+\tan B\,\left( A,\,B\neq 90^{\circ} \right)$.
Ta có $\tan A+\tan B=\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B}=\frac{\sin A\cos B+\cos A\sin B}{\cos A.\cos B}=\frac{\sin\left( A+B \right)}{\cos A.\cos B}=\frac{\sin C}{\cos A.\cos B}$.
l) $\tan A+\tan B+\tan C=\tan A.\tan B.\tan C\,\left( A,\,B,\,C\neq 90^{\circ} \right)$
Ta có $-\tan C=\tan\left( A+B \right)=\frac{\tan A+\tan B}{1-\tan A.\tan B}$.
Nên $-\tan C\left( 1-\tan A.\tan B \right)=\tan A+\tan B$.
Do đó $\tan A+\tan B+\tan C=\tan A.\tan B.\tan C$.
m) $\cot A.\cot B+\cot B.\cot C+\cot C.\cot A=1$.
$\cot A.\cot B+\cot B.\cot C+\cot C.\cot A=\cot B\left( \cot A+\cot C \right)+\cot C.\cot A$.
$=\frac{\cos B}{\sin B}\left( \frac{\sin\left( A+C \right)}{\sin A.\sin C} \right)+\cot C.\cot A$.
$=\frac{\cos B}{\sin A.\sin C}+\frac{\cos C.\cos A}{\sin A.\sin C}=\frac{\cos B+\frac{1}{2}\left( \cos\left( A+C \right)+\cos\left( A-C \right) \right)}{\sin A.\sin C}$.
$=\frac{\frac{1}{2}\left( \cos\left( A-C \right)-\cos\left( A+C \right) \right)}{\sin A.\sin C}=\frac{\sin A.\sin C}{\sin A.\sin C}=1$.
n) $\tan\frac{A}{2}.\tan\frac{B}{2}+\tan\frac{B}{2}.\tan\frac{C}{2}+\tan\frac{C}{2}.\tan\frac{A}{2}=1$.
$\tan\frac{A}{2}.\tan\frac{B}{2}+\tan\frac{B}{2}.\tan\frac{C}{2}+\tan\frac{C}{2}.\tan\frac{A}{2}=\tan\frac{B}{2}\left( \tan\frac{A}{2}+\tan\frac{C}{2} \right)+\tan\frac{C}{2}.\tan\frac{A}{2}$.
$=\frac{\sin\frac{B}{2}}{\cos\frac{B}{2}}\left( \frac{\sin\left( \frac{A}{2}+\frac{C}{2} \right)}{\cos\frac{A}{2}\cos\frac{C}{2}} \right)+\tan\frac{C}{2}.\tan\frac{A}{2}$.
$=\frac{\sin\frac{B}{2}}{\cos\frac{A}{2}\cos\frac{C}{2}}+\frac{\sin\frac{C}{2}.\sin\frac{A}{2}}{\cos\frac{A}{2}\cos\frac{C}{2}}$.
$=\frac{\sin\frac{B}{2}+\sin\frac{C}{2}.\sin\frac{A}{2}}{\cos\frac{A}{2}\cos\frac{C}{2}}$.
$=\frac{\sin\frac{B}{2}+\frac{1}{2}\left( \cos\left( \frac{C}{2}-\frac{A}{2} \right)-\cos\left( \frac{C}{2}+\frac{A}{2} \right) \right)}{\cos\frac{A}{2}\cos\frac{C}{2}}$. $=\frac{\frac{1}{2}\left( \cos\left( \frac{C}{2}-\frac{A}{2} \right)+\cos\left( \frac{C}{2}+\frac{A}{2} \right) \right)}{\cos\frac{A}{2}\cos\frac{C}{2}}=\frac{\cos\frac{A}{2}\cos\frac{C}{2}}{\cos\frac{A}{2}\cos\frac{C}{2}}=1$.
o) $\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}=\cot\frac{A}{2}.\cot\frac{B}{2}.\cot\frac{C}{2}$.
$\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}=\cot\frac{A}{2}.\cot\frac{B}{2}.\cot\frac{C}{2}$.
$\Leftrightarrow \cot\frac{A}{2}+\cot\frac{B}{2}=\cot\frac{A}{2}.\cot\frac{B}{2}.\cot\frac{C}{2}-\cot\frac{C}{2}$.
$\Leftrightarrow \cot\frac{A}{2}+\cot\frac{B}{2}=\left( \cot\frac{A}{2}.\cot\frac{B}{2}-1 \right)\cot\frac{C}{2}$.
$\Leftrightarrow \cot\frac{A}{2}+\cot\frac{B}{2}=\left( \frac{\cos\frac{A}{2}.\cos\frac{B}{2}}{\sin\frac{A}{2}.\sin\frac{B}{2}}-1 \right)\frac{\cos\frac{C}{2}}{\sin\frac{C}{2}}$.
$\Leftrightarrow \cot\frac{A}{2}+\cot\frac{B}{2}=\left( \frac{\cos\left( \frac{A}{2}+\frac{B}{2} \right)}{\sin\frac{A}{2}.\sin\frac{B}{2}} \right)\frac{\cos\frac{C}{2}}{\sin\frac{C}{2}}$.
$\Leftrightarrow \cot\frac{A}{2}+\cot\frac{B}{2}=\frac{\cos\frac{C}{2}}{\sin\frac{A}{2}.\sin\frac{B}{2}}$.
$\Leftrightarrow \frac{\cos\frac{A}{2}}{\sin\frac{A}{2}}+\frac{\cos\frac{B}{2}}{\sin\frac{B}{2}}=\frac{\cos\frac{C}{2}}{\sin\frac{A}{2}.\sin\frac{B}{2}}\Leftrightarrow \frac{\cos\frac{A}{2}\sin\frac{B}{2}+\cos\frac{B}{2}\sin\frac{A}{2}}{\sin\frac{A}{2}\sin\frac{B}{2}}=\frac{\cos\frac{C}{2}}{\sin\frac{A}{2}.\sin\frac{B}{2}}$.
$\Leftrightarrow \frac{\sin\left( \frac{A}{2}+\frac{B}{2} \right)}{\sin\frac{A}{2}\sin\frac{B}{2}}=\frac{\cos\frac{C}{2}}{\sin\frac{A}{2}.\sin\frac{B}{2}}$. Luôn đúng
Vậy $\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}=\cot\frac{A}{2}.\cot\frac{B}{2}.\cot\frac{C}{2}$.
p) $\cot B+\frac{\cos C}{\sin B.\cos A}=\cot C+\frac{\cos B}{\sin C.\cos A}\,\left( A\neq 90^{\circ} \right)$
$\cot B+\frac{\cos C}{\sin B.\cos A}=\cot C+\frac{\cos B}{\sin C.\cos A}\Leftrightarrow \cot B-\cot C=\frac{\cos B}{\sin C.\cos A}-\frac{\cos C}{\sin B.\cos A}$.
$\Leftrightarrow \frac{\cos B\sin C-\cos C\sin B}{\sin B\sin C}=\frac{1}{\cos A}\left( \frac{\frac{1}{2}\left( \sin 2B-\sin 2C \right)}{\sin B\sin C} \right)$.
$\Leftrightarrow \frac{\sin\left( C-B \right)}{\sin B\sin C}=\frac{1}{\cos A}\left( \frac{\frac{1}{2}\left( \sin 2B-\sin 2C \right)}{\sin B\sin C} \right)\Leftrightarrow \frac{\sin\left( C-B \right)}{\sin B\sin C}=\frac{1}{\cos A}\frac{\cos\left( B+C \right)\sin\left( B-C \right)}{\sin B\sin C}\Leftrightarrow \frac{\sin\left( C-B \right)}{\sin B\sin C}=\frac{1}{\cos A}\frac{-\cos A\sin\left( B-C \right)}{\sin B\sin C}\Leftrightarrow \frac{\sin\left( C-B \right)}{\sin B\sin C}=\frac{\sin\left( C-B \right)}{\sin B\sin C}$
Vậy $\cot B+\frac{\cos C}{\sin B.\cos A}=\cot C+\frac{\cos B}{\sin C.\cos A}\,\left( A\neq 90^{\circ} \right)$.
q) $\cos\frac{A}{2}.\cos\frac{B}{2}.\cos\frac{C}{2}=\sin\frac{A}{2}.\sin\frac{B}{2}.\cos\frac{C}{2}+\sin\frac{A}{2}.\cos\frac{B}{2}.\sin\frac{C}{2}+\cos\frac{A}{2}.\sin\frac{B}{2}.\sin\frac{C}{2}$.
Đặt $\sin\frac{A}{2}.\sin\frac{B}{2}.\cos\frac{C}{2}+\sin\frac{A}{2}.\cos\frac{B}{2}.\sin\frac{C}{2}+\cos\frac{A}{2}.\sin\frac{B}{2}.\sin\frac{C}{2}=T$.
$T=\sin\frac{A}{2}.\sin\left( \frac{B+C}{2} \right)+\cos\frac{A}{2}.\sin\frac{B}{2}.\sin\frac{C}{2}$.
$T=\cos\frac{A}{2}.\left( \sin\frac{A}{2}+\sin\frac{B}{2}.\sin\frac{C}{2} \right)$.
$T=\cos\frac{A}{2}.\left( \cos\frac{B+C}{2}+\sin\frac{B}{2}.\sin\frac{C}{2} \right)$.
$T=\cos\frac{A}{2}.\left( \cos\frac{B}{2}\cos\frac{C}{2}-\sin\frac{B}{2}.\sin\frac{C}{2}+\sin\frac{B}{2}.\sin\frac{C}{2} \right)=\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$.
r) $\sin^{2}\frac{A}{2}+\sin^{2}\frac{B}{2}+\sin^{2}\frac{C}{2}=1-2\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$
Ta có $\sin^{2}\frac{A}{2}+\sin^{2}\frac{B}{2}+\sin^{2}\frac{C}{2}=\frac{1-\cos A}{2}+\frac{1-\cos B}{2}+\sin^{2}\frac{C}{2}$.
$=1-\frac{1}{2}\left( \cos A+\cos B \right)+\sin^{2}\frac{C}{2}$ $=1-\frac{1}{2}\left( 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} \right)+\sin^{2}\frac{C}{2}=1-\left( \sin\frac{C}{2}\cos\frac{A-B}{2} \right)+\sin^{2}\frac{C}{2}$
$=1+\sin\frac{C}{2}\left( \cos\frac{A+B}{2}-\cos\frac{A-B}{2} \right)$
$=1-2\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$.