Bài tập nâng cao · Bài 11
Dạng 4. Biến đổi tổng thành tích
Bài tập nâng cao Công thức lượng giác · Bài 11
Tính giá trị các biểu thức sau:
a) $A=\sin\frac{\pi}{30}\sin\frac{7\pi}{30}\sin\frac{13\pi}{30}\sin\frac{19\pi}{30}\sin\frac{25\pi}{30}$
b) $B=16.\sin10^{\circ}.\sin30^{\circ}.\sin50^{\circ}.\sin70^{\circ}.\sin90^{\circ}$
c) $C=\cos24^{\circ}+\cos48^{\circ}-\cos84^{\circ}-\cos12^{\circ}$
d) $D=\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}$
e) $E=\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}$
f) $F=\cos\frac{\pi}{9}+\cos\frac{5\pi}{9}+\cos\frac{7\pi}{9}$
g) $G=\cos\frac{2\pi}{5}+\cos\frac{4\pi}{5}+\cos\frac{6\pi}{5}+\cos\frac{8\pi}{5}$
h) $H=\cos\frac{\pi}{11}+\cos\frac{3\pi}{11}+\cos\frac{5\pi}{11}+\cos\frac{7\pi}{11}+\cos\frac{9\pi}{11}$
a) $A=\sin\frac{\pi}{30}\sin\frac{7\pi}{30}\sin\frac{13\pi}{30}\sin\frac{19\pi}{30}\sin\frac{25\pi}{30}$
b) $B=16.\sin10^{\circ}.\sin30^{\circ}.\sin50^{\circ}.\sin70^{\circ}.\sin90^{\circ}$
c) $C=\cos24^{\circ}+\cos48^{\circ}-\cos84^{\circ}-\cos12^{\circ}$
d) $D=\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}$
e) $E=\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}$
f) $F=\cos\frac{\pi}{9}+\cos\frac{5\pi}{9}+\cos\frac{7\pi}{9}$
g) $G=\cos\frac{2\pi}{5}+\cos\frac{4\pi}{5}+\cos\frac{6\pi}{5}+\cos\frac{8\pi}{5}$
h) $H=\cos\frac{\pi}{11}+\cos\frac{3\pi}{11}+\cos\frac{5\pi}{11}+\cos\frac{7\pi}{11}+\cos\frac{9\pi}{11}$
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Lời giải
a) Ta có: $A=\sin\frac{\pi}{30}\sin\frac{7\pi}{30}\sin\frac{13\pi}{30}\sin\frac{19\pi}{30}\sin\frac{25\pi}{30}=\sin\frac{\pi}{30}\cos\frac{8\pi}{30}\cos\frac{2\pi}{30}\cos\frac{4\pi}{30}\sin\frac{5\pi}{6}$
$\Leftrightarrow 16.\cos\frac{\pi}{30}.A=16\cos\frac{\pi}{30}\sin\frac{\pi}{30}\cos\frac{8\pi}{30}\cos\frac{2\pi}{30}\cos\frac{4\pi}{30}\sin\frac{5\pi}{6}$
$\Leftrightarrow 16.\cos\frac{\pi}{30}.A=8.\sin\frac{2\pi}{30}\cos\frac{2\pi}{30}\cos\frac{4\pi}{30}\cos\frac{8\pi}{30}.\frac{1}{2}=4\sin\frac{4\pi}{30}\cos\frac{4\pi}{30}\cos\frac{8\pi}{30}.\frac{1}{2}$
$\Leftrightarrow 16.\cos\frac{\pi}{30}.A=2\sin\frac{8\pi}{30}\cos\frac{8\pi}{30}.\frac{1}{2}=\sin\frac{16\pi}{30}.\frac{1}{2}=\sin\left( \frac{15\pi}{30}+\frac{\pi}{30} \right).\frac{1}{2}=\frac{1}{2}\cos\frac{\pi}{30}\Rightarrow A=\frac{1}{32}$
b) Ta có: $B=16.\sin10^{\circ}.\sin30^{\circ}.\sin50^{\circ}.\sin70^{\circ}.\sin90^{\circ}=16.\sin10^{\circ}.\sin30^{\circ}.\cos40^{\circ}.\cos20^{\circ}$
$\begin{array}{l}\Leftrightarrow B.\cos10^{\circ}=16.\sin10^{\circ}.\cos10^{\circ}.\frac{1}{2}.\cos40^{\circ}.\cos20^{\circ}.1 \\ \Leftrightarrow B.\cos10^{\circ}=4\sin20^{\circ}.\cos20^{\circ}.\cos40^{\circ}. \\ \Leftrightarrow B.\cos10^{\circ}=2\sin40^{\circ}.\cos40^{\circ}=\sin80^{\circ}=\cos10^{\circ}\to B=1\end{array}$
c) $C=\left( \cos24^{\circ}+\cos48^{\circ} \right)-\left( \cos84^{\circ}+\cos12^{\circ} \right)=2\cos36^{\circ}.\cos12^{\circ}-2\cos48^{\circ}.\cos36^{\circ}$
$\begin{array}{l}=2\cos36^{\circ}\left( \cos12^{\circ}-\cos48^{\circ} \right)=4\cos36^{\circ}.\sin30^{\circ}.\sin18^{\circ} \\ =2(1-2\sin^{2}18^{\circ}).\sin18^{\circ}=-4\sin^{3}18^{\circ}+2\sin18^{\circ}\end{array}$
Ta có: $\cos36^{\circ}=\sin54^{\circ}\Leftrightarrow 1-2\sin^{2}18^{\circ}=3\sin18^{\circ}-4\sin^{3}18^{\circ}\Leftrightarrow 4\sin^{3}18^{\circ}-2\sin^{2}18^{\circ}-3\sin18^{\circ}+1=0$ Đặt $t=\sin18^{\circ},\ 0 < t < 1.$ Ta được phương trình: $4t^{3}-2t^{2}-3t+1=0\Leftrightarrow \left[ \begin{array}{l} t=1 \\ 4t^{2}+2t-1=0 \end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l} t=1\ (L) \\ t=\frac{-1-\sqrt{5}}{4}\ (L) \\ t=\frac{-1+\sqrt{5}}{4} \end{array} \right.$. Suy ra $\sin18^{\circ}=\frac{-1+\sqrt{5}}{4}$.
Vậy: $C=-4\sin^{3}18^{\circ}+2\sin18^{\circ}=\frac{1}{2}$
d) $D=\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}$
$\Leftrightarrow 2\sin\frac{2\pi}{7}.D=2\sin\frac{2\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{2\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{2\pi}{7}\cos\frac{6\pi}{7}$
$\begin{array}{l}\Leftrightarrow 2\sin\frac{2\pi}{7}.D=\sin\frac{4\pi}{7}+\sin\frac{6\pi}{7}-\sin\frac{2\pi}{7}+\sin\frac{8\pi}{7}-\sin\frac{4\pi}{7} \\ \Leftrightarrow 2\sin\frac{2\pi}{7}.D=\sin\frac{6\pi}{7}-\sin\frac{2\pi}{7}+\sin\left( 2\pi-\frac{6\pi}{7} \right)=\sin\frac{6\pi}{7}-\sin\frac{2\pi}{7}-\sin\frac{6\pi}{7}=-\sin\frac{2\pi}{7}\end{array}$
Vậy $D=\frac{-1}{2}$.
e) $E=\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}$
$\Leftrightarrow 2\sin\frac{\pi}{7}.E=2\sin\frac{\pi}{7}\cos\frac{\pi}{7}-2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{3\pi}{7}\Leftrightarrow 2\sin\frac{\pi}{7}.E=\sin\frac{2\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{\pi}{7}+\sin\frac{4\pi}{7}-\sin\frac{2\pi}{7}=\sin\frac{\pi}{7}$
Vậy $E=\frac{1}{2}$.
f) $F=\cos\frac{\pi}{9}+\cos\frac{5\pi}{9}+\cos\frac{7\pi}{9}=\cos\frac{\pi}{9}+2\cos\frac{6\pi}{9}\cos\frac{\pi}{9}=\cos\frac{\pi}{9}-\cos\frac{\pi}{9}=0$
g) $G=\cos\frac{2\pi}{5}+\cos\frac{4\pi}{5}+\cos\frac{6\pi}{5}+\cos\frac{8\pi}{5}$
$\Leftrightarrow 2\sin\frac{\pi}{5}.G=2\sin\frac{\pi}{5}.\left( \cos\frac{2\pi}{5}+\cos\frac{4\pi}{5}+\cos\frac{6\pi}{5}+\cos\frac{8\pi}{5} \right)$
$\Leftrightarrow 2\sin\frac{\pi}{5}.G=\sin\frac{3\pi}{5}-\sin\frac{\pi}{5}+\sin\frac{5\pi}{5}-\sin\frac{3\pi}{5}+\sin\frac{7\pi}{5}-\sin\frac{5\pi}{5}+\sin\frac{9\pi}{5}-\sin\frac{7\pi}{5}$
$\Leftrightarrow 2\sin\frac{\pi}{5}.G=-\sin\frac{\pi}{5}+\sin\frac{9\pi}{5}=-\sin\frac{\pi}{5}+\sin\left( 2\pi-\frac{\pi}{5} \right)=-2\sin\frac{\pi}{5}$
Vậy $G=-1$.
h) $H=\cos\frac{\pi}{11}+\cos\frac{3\pi}{11}+\cos\frac{5\pi}{11}+\cos\frac{7\pi}{11}+\cos\frac{9\pi}{11}$
$\Leftrightarrow 2\sin\frac{\pi}{11}.H=2\sin\frac{\pi}{11}.\left( \cos\frac{\pi}{11}+\cos\frac{3\pi}{11}+\cos\frac{5\pi}{11}+\cos\frac{7\pi}{11}+\cos\frac{9\pi}{11} \right)$
$\Leftrightarrow 2\sin\frac{\pi}{11}.H=\sin\frac{2\pi}{11}+\sin\frac{4\pi}{11}-\sin\frac{2\pi}{11}+\sin\frac{6\pi}{11}-\sin\frac{4\pi}{11}+\sin\frac{8\pi}{11}-\sin\frac{6\pi}{11}+\sin\frac{10\pi}{11}-\sin\frac{8\pi}{11}$
$\Leftrightarrow 2\sin\frac{\pi}{11}.H=\sin\frac{10\pi}{11}=\sin\frac{\pi}{11}\to H=\frac{1}{2}$. Vậy $H=\frac{1}{2}$.
$\Leftrightarrow 16.\cos\frac{\pi}{30}.A=16\cos\frac{\pi}{30}\sin\frac{\pi}{30}\cos\frac{8\pi}{30}\cos\frac{2\pi}{30}\cos\frac{4\pi}{30}\sin\frac{5\pi}{6}$
$\Leftrightarrow 16.\cos\frac{\pi}{30}.A=8.\sin\frac{2\pi}{30}\cos\frac{2\pi}{30}\cos\frac{4\pi}{30}\cos\frac{8\pi}{30}.\frac{1}{2}=4\sin\frac{4\pi}{30}\cos\frac{4\pi}{30}\cos\frac{8\pi}{30}.\frac{1}{2}$
$\Leftrightarrow 16.\cos\frac{\pi}{30}.A=2\sin\frac{8\pi}{30}\cos\frac{8\pi}{30}.\frac{1}{2}=\sin\frac{16\pi}{30}.\frac{1}{2}=\sin\left( \frac{15\pi}{30}+\frac{\pi}{30} \right).\frac{1}{2}=\frac{1}{2}\cos\frac{\pi}{30}\Rightarrow A=\frac{1}{32}$
b) Ta có: $B=16.\sin10^{\circ}.\sin30^{\circ}.\sin50^{\circ}.\sin70^{\circ}.\sin90^{\circ}=16.\sin10^{\circ}.\sin30^{\circ}.\cos40^{\circ}.\cos20^{\circ}$
$\begin{array}{l}\Leftrightarrow B.\cos10^{\circ}=16.\sin10^{\circ}.\cos10^{\circ}.\frac{1}{2}.\cos40^{\circ}.\cos20^{\circ}.1 \\ \Leftrightarrow B.\cos10^{\circ}=4\sin20^{\circ}.\cos20^{\circ}.\cos40^{\circ}. \\ \Leftrightarrow B.\cos10^{\circ}=2\sin40^{\circ}.\cos40^{\circ}=\sin80^{\circ}=\cos10^{\circ}\to B=1\end{array}$
c) $C=\left( \cos24^{\circ}+\cos48^{\circ} \right)-\left( \cos84^{\circ}+\cos12^{\circ} \right)=2\cos36^{\circ}.\cos12^{\circ}-2\cos48^{\circ}.\cos36^{\circ}$
$\begin{array}{l}=2\cos36^{\circ}\left( \cos12^{\circ}-\cos48^{\circ} \right)=4\cos36^{\circ}.\sin30^{\circ}.\sin18^{\circ} \\ =2(1-2\sin^{2}18^{\circ}).\sin18^{\circ}=-4\sin^{3}18^{\circ}+2\sin18^{\circ}\end{array}$
Ta có: $\cos36^{\circ}=\sin54^{\circ}\Leftrightarrow 1-2\sin^{2}18^{\circ}=3\sin18^{\circ}-4\sin^{3}18^{\circ}\Leftrightarrow 4\sin^{3}18^{\circ}-2\sin^{2}18^{\circ}-3\sin18^{\circ}+1=0$ Đặt $t=\sin18^{\circ},\ 0 < t < 1.$ Ta được phương trình: $4t^{3}-2t^{2}-3t+1=0\Leftrightarrow \left[ \begin{array}{l} t=1 \\ 4t^{2}+2t-1=0 \end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l} t=1\ (L) \\ t=\frac{-1-\sqrt{5}}{4}\ (L) \\ t=\frac{-1+\sqrt{5}}{4} \end{array} \right.$. Suy ra $\sin18^{\circ}=\frac{-1+\sqrt{5}}{4}$.
Vậy: $C=-4\sin^{3}18^{\circ}+2\sin18^{\circ}=\frac{1}{2}$
d) $D=\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}$
$\Leftrightarrow 2\sin\frac{2\pi}{7}.D=2\sin\frac{2\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{2\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{2\pi}{7}\cos\frac{6\pi}{7}$
$\begin{array}{l}\Leftrightarrow 2\sin\frac{2\pi}{7}.D=\sin\frac{4\pi}{7}+\sin\frac{6\pi}{7}-\sin\frac{2\pi}{7}+\sin\frac{8\pi}{7}-\sin\frac{4\pi}{7} \\ \Leftrightarrow 2\sin\frac{2\pi}{7}.D=\sin\frac{6\pi}{7}-\sin\frac{2\pi}{7}+\sin\left( 2\pi-\frac{6\pi}{7} \right)=\sin\frac{6\pi}{7}-\sin\frac{2\pi}{7}-\sin\frac{6\pi}{7}=-\sin\frac{2\pi}{7}\end{array}$
Vậy $D=\frac{-1}{2}$.
e) $E=\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}$
$\Leftrightarrow 2\sin\frac{\pi}{7}.E=2\sin\frac{\pi}{7}\cos\frac{\pi}{7}-2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{3\pi}{7}\Leftrightarrow 2\sin\frac{\pi}{7}.E=\sin\frac{2\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{\pi}{7}+\sin\frac{4\pi}{7}-\sin\frac{2\pi}{7}=\sin\frac{\pi}{7}$
Vậy $E=\frac{1}{2}$.
f) $F=\cos\frac{\pi}{9}+\cos\frac{5\pi}{9}+\cos\frac{7\pi}{9}=\cos\frac{\pi}{9}+2\cos\frac{6\pi}{9}\cos\frac{\pi}{9}=\cos\frac{\pi}{9}-\cos\frac{\pi}{9}=0$
g) $G=\cos\frac{2\pi}{5}+\cos\frac{4\pi}{5}+\cos\frac{6\pi}{5}+\cos\frac{8\pi}{5}$
$\Leftrightarrow 2\sin\frac{\pi}{5}.G=2\sin\frac{\pi}{5}.\left( \cos\frac{2\pi}{5}+\cos\frac{4\pi}{5}+\cos\frac{6\pi}{5}+\cos\frac{8\pi}{5} \right)$
$\Leftrightarrow 2\sin\frac{\pi}{5}.G=\sin\frac{3\pi}{5}-\sin\frac{\pi}{5}+\sin\frac{5\pi}{5}-\sin\frac{3\pi}{5}+\sin\frac{7\pi}{5}-\sin\frac{5\pi}{5}+\sin\frac{9\pi}{5}-\sin\frac{7\pi}{5}$
$\Leftrightarrow 2\sin\frac{\pi}{5}.G=-\sin\frac{\pi}{5}+\sin\frac{9\pi}{5}=-\sin\frac{\pi}{5}+\sin\left( 2\pi-\frac{\pi}{5} \right)=-2\sin\frac{\pi}{5}$
Vậy $G=-1$.
h) $H=\cos\frac{\pi}{11}+\cos\frac{3\pi}{11}+\cos\frac{5\pi}{11}+\cos\frac{7\pi}{11}+\cos\frac{9\pi}{11}$
$\Leftrightarrow 2\sin\frac{\pi}{11}.H=2\sin\frac{\pi}{11}.\left( \cos\frac{\pi}{11}+\cos\frac{3\pi}{11}+\cos\frac{5\pi}{11}+\cos\frac{7\pi}{11}+\cos\frac{9\pi}{11} \right)$
$\Leftrightarrow 2\sin\frac{\pi}{11}.H=\sin\frac{2\pi}{11}+\sin\frac{4\pi}{11}-\sin\frac{2\pi}{11}+\sin\frac{6\pi}{11}-\sin\frac{4\pi}{11}+\sin\frac{8\pi}{11}-\sin\frac{6\pi}{11}+\sin\frac{10\pi}{11}-\sin\frac{8\pi}{11}$
$\Leftrightarrow 2\sin\frac{\pi}{11}.H=\sin\frac{10\pi}{11}=\sin\frac{\pi}{11}\to H=\frac{1}{2}$. Vậy $H=\frac{1}{2}$.