Bài tập nâng cao · Bài 5
Dạng 2. Công thức nhân đôi
Bài tập nâng cao Công thức lượng giác · Bài 5
Tính giá trị biểu thức:
a. $A=\sin\frac{\pi}{8}\cos\frac{\pi}{4}\cos\frac{\pi}{8}$
b. $B=\frac{1-\tan^{2}\frac{\pi}{8}}{\tan\frac{\pi}{8}}$
c. $C=\sin 10^{\circ}\sin 50^{\circ}\sin 70^{\circ}$
d. $D=\sin 6^{\circ}\sin 42^{\circ}\sin 66^{\circ}\sin 78^{\circ}$
e. $E=16\cos 20^{\circ}\cos 40^{\circ}\cos 60^{\circ}\cos 80^{\circ}$
a. $A=\sin\frac{\pi}{8}\cos\frac{\pi}{4}\cos\frac{\pi}{8}$
b. $B=\frac{1-\tan^{2}\frac{\pi}{8}}{\tan\frac{\pi}{8}}$
c. $C=\sin 10^{\circ}\sin 50^{\circ}\sin 70^{\circ}$
d. $D=\sin 6^{\circ}\sin 42^{\circ}\sin 66^{\circ}\sin 78^{\circ}$
e. $E=16\cos 20^{\circ}\cos 40^{\circ}\cos 60^{\circ}\cos 80^{\circ}$
Xem lời giải
Lời giải
Áp dụng công thức: $\sin 2a=2\sin a.\cos a$ ta có:
$A=\sin\frac{\pi}{8}\cos\frac{\pi}{4}\cos\frac{\pi}{8}=\sin\frac{\pi}{8}\cos\frac{\pi}{8}\cos\frac{\pi}{4}=\frac{1}{2}\sin\frac{\pi}{4}\cos\frac{\pi}{4}=\frac{1}{4}\sin\frac{\pi}{2}=\frac{1}{4}$
b. Áp dụng công thức: $\tan 2a=\frac{2\tan a}{1-\tan^{2}a}\Rightarrow \frac{1-\tan^{2}a}{2\tan a}=\frac{1}{\tan 2a}=\cot 2a$
ta có: $B=\frac{1-\tan^{2}\frac{\pi}{8}}{\tan\frac{\pi}{8}}=2.\frac{1-\tan^{2}\frac{\pi}{8}}{2\tan\frac{\pi}{8}}=2\cot(2.\frac{\pi}{8})=2\cot\frac{\pi}{4}=2$
c. Áp dụng công thức: $\sin\left( \frac{\pi}{2}-a \right)=\cos a$ và $\sin 2a=2\sin a.\cos a$ ta có:
$\begin{array}{l}C=\sin 10^{\circ}\sin 50^{\circ}\sin 70^{\circ}=\sin 10^{\circ}.\cos 40^{\circ}\cos 20^{\circ}=\frac{2\sin 10^{\circ}\cos 10^{\circ}.\cos 20^{\circ}\cos 40^{\circ}}{2\cos 10^{\circ}}\\=\frac{\sin 20^{\circ}.\cos 20^{\circ}\cos 40^{\circ}}{2\cos 10^{\circ}}=\frac{\sin 40^{\circ}\cos 40^{\circ}}{4\cos 10^{\circ}}=\frac{\sin 80^{\circ}}{8\cos 10^{\circ}}=\frac{\cos 10^{\circ}}{8\cos 10^{\circ}}=\frac{1}{8}\end{array}$
d. Tương tự câu c ta có:
$\begin{array}{l}D=\sin 6^{\circ}\sin 42^{\circ}\sin 66^{\circ}\sin 78^{\circ}=\sin 6^{\circ}\cos 48^{\circ}\cos 24^{\circ}\cos 12^{\circ}\\=\frac{2\sin 6^{\circ}\cos 6^{\circ}\cos 12^{\circ}\cos 24^{\circ}\cos 48^{\circ}}{2\cos 6^{\circ}}\\=\frac{\sin 12^{\circ}\cos 12^{\circ}\cos 24^{\circ}\cos 48^{\circ}}{2\cos 6^{\circ}}\\=\frac{\sin 24^{\circ}\cos 24^{\circ}\cos 48^{\circ}}{4\cos 6^{\circ}}=\frac{\sin 48^{\circ}\cos 48^{\circ}}{8\cos 6^{\circ}}=\frac{\sin 96^{\circ}}{16\cos 6^{\circ}}\end{array}$
Do: $\sin 96^{\circ}=\sin\left( 90^{\circ}-\left( -6^{\circ} \right) \right)=\cos\left( -6^{\circ} \right)=\cos 6^{\circ}$ nên $D=\frac{1}{16}$
e. Ta có: $\begin{array}{l}E=16\cos 20^{\circ}\cos 40^{\circ}\cos 60^{\circ}\cos 80^{\circ}=\frac{8\sin 20^{\circ}\cos 20^{\circ}\cos 40^{\circ}\cos 80^{\circ}}{\sin 20^{\circ}}\\=\frac{4\sin 40^{\circ}\cos 40^{\circ}\cos 80^{\circ}}{\sin 20^{\circ}}=\frac{2\sin 80^{\circ}\cos 80^{\circ}}{\sin 20^{\circ}}=\frac{\sin 160^{\circ}}{\sin 20^{\circ}}=1\end{array}$
$A=\sin\frac{\pi}{8}\cos\frac{\pi}{4}\cos\frac{\pi}{8}=\sin\frac{\pi}{8}\cos\frac{\pi}{8}\cos\frac{\pi}{4}=\frac{1}{2}\sin\frac{\pi}{4}\cos\frac{\pi}{4}=\frac{1}{4}\sin\frac{\pi}{2}=\frac{1}{4}$
b. Áp dụng công thức: $\tan 2a=\frac{2\tan a}{1-\tan^{2}a}\Rightarrow \frac{1-\tan^{2}a}{2\tan a}=\frac{1}{\tan 2a}=\cot 2a$
ta có: $B=\frac{1-\tan^{2}\frac{\pi}{8}}{\tan\frac{\pi}{8}}=2.\frac{1-\tan^{2}\frac{\pi}{8}}{2\tan\frac{\pi}{8}}=2\cot(2.\frac{\pi}{8})=2\cot\frac{\pi}{4}=2$
c. Áp dụng công thức: $\sin\left( \frac{\pi}{2}-a \right)=\cos a$ và $\sin 2a=2\sin a.\cos a$ ta có:
$\begin{array}{l}C=\sin 10^{\circ}\sin 50^{\circ}\sin 70^{\circ}=\sin 10^{\circ}.\cos 40^{\circ}\cos 20^{\circ}=\frac{2\sin 10^{\circ}\cos 10^{\circ}.\cos 20^{\circ}\cos 40^{\circ}}{2\cos 10^{\circ}}\\=\frac{\sin 20^{\circ}.\cos 20^{\circ}\cos 40^{\circ}}{2\cos 10^{\circ}}=\frac{\sin 40^{\circ}\cos 40^{\circ}}{4\cos 10^{\circ}}=\frac{\sin 80^{\circ}}{8\cos 10^{\circ}}=\frac{\cos 10^{\circ}}{8\cos 10^{\circ}}=\frac{1}{8}\end{array}$
d. Tương tự câu c ta có:
$\begin{array}{l}D=\sin 6^{\circ}\sin 42^{\circ}\sin 66^{\circ}\sin 78^{\circ}=\sin 6^{\circ}\cos 48^{\circ}\cos 24^{\circ}\cos 12^{\circ}\\=\frac{2\sin 6^{\circ}\cos 6^{\circ}\cos 12^{\circ}\cos 24^{\circ}\cos 48^{\circ}}{2\cos 6^{\circ}}\\=\frac{\sin 12^{\circ}\cos 12^{\circ}\cos 24^{\circ}\cos 48^{\circ}}{2\cos 6^{\circ}}\\=\frac{\sin 24^{\circ}\cos 24^{\circ}\cos 48^{\circ}}{4\cos 6^{\circ}}=\frac{\sin 48^{\circ}\cos 48^{\circ}}{8\cos 6^{\circ}}=\frac{\sin 96^{\circ}}{16\cos 6^{\circ}}\end{array}$
Do: $\sin 96^{\circ}=\sin\left( 90^{\circ}-\left( -6^{\circ} \right) \right)=\cos\left( -6^{\circ} \right)=\cos 6^{\circ}$ nên $D=\frac{1}{16}$
e. Ta có: $\begin{array}{l}E=16\cos 20^{\circ}\cos 40^{\circ}\cos 60^{\circ}\cos 80^{\circ}=\frac{8\sin 20^{\circ}\cos 20^{\circ}\cos 40^{\circ}\cos 80^{\circ}}{\sin 20^{\circ}}\\=\frac{4\sin 40^{\circ}\cos 40^{\circ}\cos 80^{\circ}}{\sin 20^{\circ}}=\frac{2\sin 80^{\circ}\cos 80^{\circ}}{\sin 20^{\circ}}=\frac{\sin 160^{\circ}}{\sin 20^{\circ}}=1\end{array}$