Bài tập nâng cao · Bài 15
Dạng 4. Biến đổi tổng thành tích
Bài tập nâng cao Công thức lượng giác · Bài 15
Chứng minh các đẳng thức lượng giác:
a. $\tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ}=4$
b. $\tan 20^{\circ}-\tan 40^{\circ}+\tan 80^{\circ}=3\sqrt{3}$
c. $\tan 10^{\circ}-\tan 50^{\circ}+\tan 60^{\circ}+\tan 70^{\circ}=2\sqrt{3}$
d. $\tan 30^{\circ}+\tan 40^{\circ}+\tan 50^{\circ}+\tan 60^{\circ}=\frac{8\sqrt{3}}{3}.\cos 20^{\circ}$
e. $\tan^{6}20^{\circ}-33\tan^{4}20^{\circ}+27\tan^{2}20^{\circ}-3=0$
a. $\tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ}=4$
b. $\tan 20^{\circ}-\tan 40^{\circ}+\tan 80^{\circ}=3\sqrt{3}$
c. $\tan 10^{\circ}-\tan 50^{\circ}+\tan 60^{\circ}+\tan 70^{\circ}=2\sqrt{3}$
d. $\tan 30^{\circ}+\tan 40^{\circ}+\tan 50^{\circ}+\tan 60^{\circ}=\frac{8\sqrt{3}}{3}.\cos 20^{\circ}$
e. $\tan^{6}20^{\circ}-33\tan^{4}20^{\circ}+27\tan^{2}20^{\circ}-3=0$
Xem lời giải
Lời giải
a. $VT=\frac{\sin 9^{\circ}}{\cos 9^{\circ}}+\frac{\sin 81^{\circ}}{\cos 81^{\circ}}-\left( \frac{\sin 27^{\circ}}{\cos 27^{\circ}}+\frac{\sin 63^{\circ}}{\cos 63^{\circ}} \right)$
$\begin{array}{l} =\frac{\sin 9^{\circ}\cos 81^{\circ}+\cos 9^{\circ}\sin 81^{\circ}}{\cos 9^{\circ}\cos 81^{\circ}}-\frac{\sin 27^{\circ}\cos 63^{\circ}+\cos 27^{\circ}\sin 63^{\circ}}{\cos 27^{\circ}\cos 63^{\circ}} \\ =\frac{\sin\left( 9^{\circ}+81^{\circ} \right)}{\cos 9^{\circ}\sin 9^{\circ}}-\frac{\sin\left( 27^{\circ}+63^{\circ} \right)}{\cos 27^{\circ}\sin 27^{\circ}}=\frac{2}{\sin 18^{\circ}}-\frac{2}{\sin 54^{\circ}}=\frac{2\left( \sin 54^{\circ}-\sin 18^{\circ} \right)}{\sin 18^{\circ}\sin 54^{\circ}}=\frac{2.2\cos 36^{\circ}\sin 18^{\circ}}{\sin 18^{\circ}\sin 54^{\circ}}=4 \end{array}$
b. $VT=\tan 20^{\circ}-\tan\left( 60^{\circ}-20^{\circ} \right)+\tan\left( 60^{\circ}+20^{\circ} \right)$
$\begin{array}{l} =\tan 20^{\circ}-\frac{\tan 60^{\circ}-\tan 20^{\circ}}{1+\tan 60^{\circ}\tan 20^{\circ}}+\frac{\tan 60^{\circ}+\tan 20^{\circ}}{1-\tan 60^{\circ}\tan 20^{\circ}} \\ =\tan 20^{\circ}-\frac{\sqrt{3}-\tan 20^{\circ}}{1+\sqrt{3}\tan 20^{\circ}}+\frac{\sqrt{3}+\tan 20^{\circ}}{1-\sqrt{3}\tan 20^{\circ}} \\ =\tan 20^{\circ}+\frac{\left( \sqrt{3}+\tan 20^{\circ} \right)\left( 1+\sqrt{3}\tan 20^{\circ} \right)-\left( \sqrt{3}-\tan 20^{\circ} \right)\left( 1-\sqrt{3}\tan 20^{\circ} \right)}{1-3\tan^{2}20^{\circ}} \end{array}$
$=\tan 20^{\circ}+\frac{8\tan 20^{\circ}}{1-3\tan^{2}20^{\circ}}=\frac{9\tan 20^{\circ}-3\tan^{3}20^{\circ}}{1-3\tan^{2}20^{\circ}}$ $=\frac{3\tan 20^{\circ}\left( 3-\tan^{2}20^{\circ} \right)}{1-3\tan^{2}20^{\circ}}$
$=3\tan 60^{\circ}=3\sqrt{3}$ (công thức nhân ba)
* Từ câu này ta chứng minh được công thức tổng quát:
$\tan a-\tan\left( 60^{\circ}-a \right)+\tan\left( 60^{\circ}+a \right)=3\tan 3a$
c. Chứng minh tương tự câu b ta có
$\tan 10^{\circ}-\tan 50^{\circ}+\tan 70^{\circ}=\tan 10^{\circ}-\tan\left( 60^{\circ}-10^{\circ} \right)+\tan\left( 60^{\circ}+10^{\circ} \right)=3\tan 30^{\circ}=3.\frac{\sqrt{3}}{3}=\sqrt{3}$ $\Rightarrow \tan 10^{\circ}-\tan 50^{\circ}+\tan 60^{\circ}+\tan 70^{\circ}=\sqrt{3}+\sqrt{3}=2\sqrt{3}$
d. $VT=\frac{\sin 30^{\circ}}{\cos 30^{\circ}}+\frac{\sin 40^{\circ}}{\cos 40^{\circ}}+\frac{\sin 50^{\circ}}{\cos 50^{\circ}}+\frac{\sin 60^{\circ}}{\cos 60^{\circ}}$
$\begin{array}{l} =\frac{\sin 30^{\circ}\cos 60^{\circ}+\cos 30^{\circ}\sin 60^{\circ}}{\cos 30^{\circ}\cos 60^{\circ}}+\frac{\sin 40^{\circ}\cos 50^{\circ}+\cos 40^{\circ}\sin 50^{\circ}}{\cos 40^{\circ}\cos 50^{\circ}} \\ =\frac{\sin\left( 30^{\circ}+60^{\circ} \right)}{\cos 30^{\circ}\sin 30^{\circ}}+\frac{\sin\left( 40^{\circ}+50^{\circ} \right)}{\cos 40^{\circ}\sin 40^{\circ}}=\frac{2}{\sin 60^{\circ}}+\frac{2}{\sin 80^{\circ}}=\frac{2\left( \sin 80^{\circ}+\sin 60^{\circ} \right)}{\sin 60^{\circ}\sin 80^{\circ}} \\ =\frac{2.2\sin 70^{\circ}\cos 10^{\circ}}{\frac{\sqrt{3}}{2}\cos 10^{\circ}}=\frac{8\sqrt{3}}{3}\sin 70^{\circ}=\frac{8\sqrt{3}}{3}\cos 20^{\circ} \end{array}$
e. $\tan^{6}20^{\circ}-33\tan^{4}20^{\circ}+27\tan^{2}20^{\circ}-3=0$ $\begin{array}{l} \Leftrightarrow \tan^{6}20^{\circ}-33\tan^{4}20^{\circ}+27\tan^{2}20^{\circ}=3 \\ \Leftrightarrow \tan^{6}20^{\circ}-6\tan^{4}20^{\circ}+9\tan^{2}20^{\circ}=27\tan^{4}20^{\circ}-18\tan^{2}20^{\circ}+3 \\ \Leftrightarrow \left( \tan^{3}20^{\circ}-3\tan 20^{\circ} \right)^{2}=3\left( 1-3\tan^{2}20^{\circ} \right)^{2}\Leftrightarrow \left( \frac{\tan^{3}20^{\circ}-3\tan 20^{\circ}}{1-3\tan^{2}20^{\circ}} \right)^{2}=3 \end{array}$
$\Leftrightarrow \left( \tan\left( 20^{\circ}.3 \right) \right)^{2}=3$ $\Leftrightarrow \left( \tan 60^{\circ} \right)^{2}=3$ $\Leftrightarrow \left( \sqrt{3} \right)^{2}=3$ (luôn đúng)
$\begin{array}{l} =\frac{\sin 9^{\circ}\cos 81^{\circ}+\cos 9^{\circ}\sin 81^{\circ}}{\cos 9^{\circ}\cos 81^{\circ}}-\frac{\sin 27^{\circ}\cos 63^{\circ}+\cos 27^{\circ}\sin 63^{\circ}}{\cos 27^{\circ}\cos 63^{\circ}} \\ =\frac{\sin\left( 9^{\circ}+81^{\circ} \right)}{\cos 9^{\circ}\sin 9^{\circ}}-\frac{\sin\left( 27^{\circ}+63^{\circ} \right)}{\cos 27^{\circ}\sin 27^{\circ}}=\frac{2}{\sin 18^{\circ}}-\frac{2}{\sin 54^{\circ}}=\frac{2\left( \sin 54^{\circ}-\sin 18^{\circ} \right)}{\sin 18^{\circ}\sin 54^{\circ}}=\frac{2.2\cos 36^{\circ}\sin 18^{\circ}}{\sin 18^{\circ}\sin 54^{\circ}}=4 \end{array}$
b. $VT=\tan 20^{\circ}-\tan\left( 60^{\circ}-20^{\circ} \right)+\tan\left( 60^{\circ}+20^{\circ} \right)$
$\begin{array}{l} =\tan 20^{\circ}-\frac{\tan 60^{\circ}-\tan 20^{\circ}}{1+\tan 60^{\circ}\tan 20^{\circ}}+\frac{\tan 60^{\circ}+\tan 20^{\circ}}{1-\tan 60^{\circ}\tan 20^{\circ}} \\ =\tan 20^{\circ}-\frac{\sqrt{3}-\tan 20^{\circ}}{1+\sqrt{3}\tan 20^{\circ}}+\frac{\sqrt{3}+\tan 20^{\circ}}{1-\sqrt{3}\tan 20^{\circ}} \\ =\tan 20^{\circ}+\frac{\left( \sqrt{3}+\tan 20^{\circ} \right)\left( 1+\sqrt{3}\tan 20^{\circ} \right)-\left( \sqrt{3}-\tan 20^{\circ} \right)\left( 1-\sqrt{3}\tan 20^{\circ} \right)}{1-3\tan^{2}20^{\circ}} \end{array}$
$=\tan 20^{\circ}+\frac{8\tan 20^{\circ}}{1-3\tan^{2}20^{\circ}}=\frac{9\tan 20^{\circ}-3\tan^{3}20^{\circ}}{1-3\tan^{2}20^{\circ}}$ $=\frac{3\tan 20^{\circ}\left( 3-\tan^{2}20^{\circ} \right)}{1-3\tan^{2}20^{\circ}}$
$=3\tan 60^{\circ}=3\sqrt{3}$ (công thức nhân ba)
* Từ câu này ta chứng minh được công thức tổng quát:
$\tan a-\tan\left( 60^{\circ}-a \right)+\tan\left( 60^{\circ}+a \right)=3\tan 3a$
c. Chứng minh tương tự câu b ta có
$\tan 10^{\circ}-\tan 50^{\circ}+\tan 70^{\circ}=\tan 10^{\circ}-\tan\left( 60^{\circ}-10^{\circ} \right)+\tan\left( 60^{\circ}+10^{\circ} \right)=3\tan 30^{\circ}=3.\frac{\sqrt{3}}{3}=\sqrt{3}$ $\Rightarrow \tan 10^{\circ}-\tan 50^{\circ}+\tan 60^{\circ}+\tan 70^{\circ}=\sqrt{3}+\sqrt{3}=2\sqrt{3}$
d. $VT=\frac{\sin 30^{\circ}}{\cos 30^{\circ}}+\frac{\sin 40^{\circ}}{\cos 40^{\circ}}+\frac{\sin 50^{\circ}}{\cos 50^{\circ}}+\frac{\sin 60^{\circ}}{\cos 60^{\circ}}$
$\begin{array}{l} =\frac{\sin 30^{\circ}\cos 60^{\circ}+\cos 30^{\circ}\sin 60^{\circ}}{\cos 30^{\circ}\cos 60^{\circ}}+\frac{\sin 40^{\circ}\cos 50^{\circ}+\cos 40^{\circ}\sin 50^{\circ}}{\cos 40^{\circ}\cos 50^{\circ}} \\ =\frac{\sin\left( 30^{\circ}+60^{\circ} \right)}{\cos 30^{\circ}\sin 30^{\circ}}+\frac{\sin\left( 40^{\circ}+50^{\circ} \right)}{\cos 40^{\circ}\sin 40^{\circ}}=\frac{2}{\sin 60^{\circ}}+\frac{2}{\sin 80^{\circ}}=\frac{2\left( \sin 80^{\circ}+\sin 60^{\circ} \right)}{\sin 60^{\circ}\sin 80^{\circ}} \\ =\frac{2.2\sin 70^{\circ}\cos 10^{\circ}}{\frac{\sqrt{3}}{2}\cos 10^{\circ}}=\frac{8\sqrt{3}}{3}\sin 70^{\circ}=\frac{8\sqrt{3}}{3}\cos 20^{\circ} \end{array}$
e. $\tan^{6}20^{\circ}-33\tan^{4}20^{\circ}+27\tan^{2}20^{\circ}-3=0$ $\begin{array}{l} \Leftrightarrow \tan^{6}20^{\circ}-33\tan^{4}20^{\circ}+27\tan^{2}20^{\circ}=3 \\ \Leftrightarrow \tan^{6}20^{\circ}-6\tan^{4}20^{\circ}+9\tan^{2}20^{\circ}=27\tan^{4}20^{\circ}-18\tan^{2}20^{\circ}+3 \\ \Leftrightarrow \left( \tan^{3}20^{\circ}-3\tan 20^{\circ} \right)^{2}=3\left( 1-3\tan^{2}20^{\circ} \right)^{2}\Leftrightarrow \left( \frac{\tan^{3}20^{\circ}-3\tan 20^{\circ}}{1-3\tan^{2}20^{\circ}} \right)^{2}=3 \end{array}$
$\Leftrightarrow \left( \tan\left( 20^{\circ}.3 \right) \right)^{2}=3$ $\Leftrightarrow \left( \tan 60^{\circ} \right)^{2}=3$ $\Leftrightarrow \left( \sqrt{3} \right)^{2}=3$ (luôn đúng)