Bài tập nâng cao · Bài 1
Dạng 1. Công thức cộng
Bài tập nâng cao Công thức lượng giác · Bài 1
Tính giá trị của biểu thức
a.$A=\sin^{2}20^{\circ}+\sin^{2}100^{\circ}+\sin^{2}140^{\circ}$
b.$B=\cos^{2}10^{\circ}+\cos^{2}110^{\circ}+\cos^{2}130^{\circ}$
c.$C=\tan 20^{\circ}.\tan 80^{\circ}+\tan 80^{\circ}.\tan 140^{\circ}+\tan 140^{\circ}.\tan 20^{\circ}$
d.$D=\tan 10^{\circ}.\tan 70^{\circ}+\tan 70^{\circ}.\tan 130^{\circ}+\tan 130^{\circ}.\tan 190^{\circ}$
e.$E=\frac{\cot 225^{\circ}-\cot 79^{\circ}.\cot 71^{\circ}}{\cot 259^{\circ}+\cot 251^{\circ}}$
f.$F=\cos^{2}75^{\circ}-\sin^{2}75^{\circ}$
g.$G=\frac{1-\tan 15^{\circ}}{1+\tan 15^{\circ}}$
h.$H=\tan 15^{\circ}+\cot 15^{\circ}$.
a.$A=\sin^{2}20^{\circ}+\sin^{2}100^{\circ}+\sin^{2}140^{\circ}$
b.$B=\cos^{2}10^{\circ}+\cos^{2}110^{\circ}+\cos^{2}130^{\circ}$
c.$C=\tan 20^{\circ}.\tan 80^{\circ}+\tan 80^{\circ}.\tan 140^{\circ}+\tan 140^{\circ}.\tan 20^{\circ}$
d.$D=\tan 10^{\circ}.\tan 70^{\circ}+\tan 70^{\circ}.\tan 130^{\circ}+\tan 130^{\circ}.\tan 190^{\circ}$
e.$E=\frac{\cot 225^{\circ}-\cot 79^{\circ}.\cot 71^{\circ}}{\cot 259^{\circ}+\cot 251^{\circ}}$
f.$F=\cos^{2}75^{\circ}-\sin^{2}75^{\circ}$
g.$G=\frac{1-\tan 15^{\circ}}{1+\tan 15^{\circ}}$
h.$H=\tan 15^{\circ}+\cot 15^{\circ}$.
Xem lời giải
Lời giải
a.Tính $A=\sin^{2}20^{\circ}+\sin^{2}100^{\circ}+\sin^{2}140^{\circ}$
$\begin{array}{l}A=\sin^{2}20^{\circ}+\sin^{2}100^{\circ}+\sin^{2}140^{\circ}\\=\sin^{2}20^{\circ}+\sin^{2}80^{\circ}+\sin^{2}40^{\circ}\\=\frac{1-\cos 40^{\circ}}{2}+\frac{1-\cos 160^{\circ}}{2}+\frac{1-\cos 80^{\circ}}{2}\\=\frac{3}{2}-\frac{\cos 40^{\circ}+\cos 160^{\circ}+\cos 80^{\circ}}{2}\end{array}$
Mà ta có:
$\begin{array}{l}\cos 40^{\circ}+\cos 160^{\circ}+\cos 80^{\circ}=\left( \cos 40^{\circ}+\cos 80^{\circ} \right)+\cos 160^{\circ}\\=2\cos 60^{\circ}\cos 20^{\circ}-\cos 20^{\circ}\\=\cos 20^{\circ}\left( 2\cos 60^{\circ}-1 \right)=0\end{array}$
Vậy $A=\frac{3}{2}$.
b.Tính $B=\cos^{2}10^{\circ}+\cos^{2}110^{\circ}+\cos^{2}130^{\circ}$
$\begin{array}{l}B=\cos^{2}10^{\circ}+\cos^{2}110^{\circ}+\cos^{2}130^{\circ}=\cos^{2}10^{\circ}+\cos^{2}70^{\circ}+\cos^{2}50^{\circ}\\=\frac{1+\cos 20^{\circ}}{2}+\frac{1+\cos 140^{\circ}}{2}+\frac{1+\cos 100^{\circ}}{2}\end{array}$
$\begin{array}{l}=\frac{3}{2}+\frac{\left( \cos 20^{\circ}+\cos 100^{\circ} \right)+\cos 140^{\circ}}{2}\\=\frac{3}{2}+\frac{2\cos 40^{\circ}.\cos 60^{\circ}-\cos 40^{\circ}}{2}\\=\frac{3}{2}+\frac{\cos 40^{\circ}\left( 2\cos 60^{\circ}-1 \right)}{2}=\frac{3}{2}\end{array}$
c.Tính $C=\tan 20^{\circ}.\tan 80^{\circ}+\tan 80^{\circ}.\tan 140^{\circ}+\tan 140^{\circ}.\tan 20^{\circ}$
Ta chứng minh công thức sau
$\tan x.\tan\left( x+\frac{\pi}{3} \right)+\tan\left( x+\frac{\pi}{3} \right).\tan\left( x+\frac{2\pi}{3} \right)+\tan x.\tan\left( x+\frac{2\pi}{3} \right)=-3$
Nhận Xét:$\tan\left( a-b \right)=\frac{\tan a-\tan b}{1+\tan a.\tan b}\Rightarrow \tan a.\tan b=\frac{\tan a-\tan b}{\tan\left( a-b \right)}-1$
Do vậy:
$\tan x.\tan\left( x+\frac{\pi}{3} \right)=\frac{\tan\left( x+\frac{\pi}{3} \right)-\tan x}{\tan\frac{\pi}{3}}-1=\frac{1}{\sqrt{3}}\left[ \tan\left( x+\frac{\pi}{3} \right)-\tan x \right]-1$ $\left( * \right)$
$\tan\left( x+\frac{\pi}{3} \right).\tan\left( x+\frac{2\pi}{3} \right)=\frac{\tan\left( x+\frac{2\pi}{3} \right)-\tan\left( x+\frac{\pi}{3} \right)}{\tan\frac{\pi}{3}}-1$
$=\frac{1}{\sqrt{3}}\left[ \tan\left( x+\frac{2\pi}{3} \right)-\tan\left( x+\frac{\pi}{3} \right) \right]-1$ $\left( ** \right)$
$\tan\left( x+\frac{2\pi}{3} \right).\tan\left( x \right)=\frac{\tan\left( x+\frac{2\pi}{3} \right)-\tan x}{\tan\frac{2\pi}{3}}-1=-\frac{1}{\sqrt{3}}\left[ \tan\left( x+\frac{2\pi}{3} \right)-\tan x \right]-1$ $\left( *** \right)$
Cộng theo vế $\left( * \right)$ $\left( ** \right)$ $\left( *** \right)$ ta được:
$\tan x.\tan\left( x+\frac{\pi}{3} \right)+\tan\left( x+\frac{\pi}{3} \right).\tan\left( x+\frac{2\pi}{3} \right)+\tan x.\tan\left( x+\frac{2\pi}{3} \right)=-3$
Vậy $C=\tan 20^{\circ}.\tan 80^{\circ}+\tan 80^{\circ}.\tan 140^{\circ}+\tan 140^{\circ}.\tan 20^{\circ}=-3$
d.Tương tự câu c
e.$E=\frac{\cot 225^{\circ}-\cot 79^{\circ}.\cot 71^{\circ}}{\cot 259^{\circ}+\cot 251^{\circ}}$
$\cot 225^{\circ}=\cot\left( 180^{\circ}+45^{\circ} \right)=\cot 45^{\circ}$
$\cot 79^{\circ}=\tan 11^{\circ}$
$\cot 71^{\circ}=\tan 19^{\circ}$
$\cot 259^{\circ}=\cot\left( 180^{\circ}+79^{\circ} \right)=\cot 79^{\circ}=\tan 11^{\circ}$
$\cot 251^{\circ}=\cot\left( 180^{\circ}+71^{\circ} \right)=\cot\left( 71^{\circ} \right)=\tan 19^{\circ}$
Vậy $E=\frac{\cot 225^{\circ}-\cot 79^{\circ}.\cot 71^{\circ}}{\cot 259^{\circ}+\cot 251^{\circ}}=\frac{1-\tan 11^{\circ}.\tan 19^{\circ}}{\tan 11^{\circ}+\tan 19^{\circ}}=\cot\left( 11^{\circ}+19^{\circ} \right)=\cot 30^{\circ}=\sqrt{3}$
f.$F=\cos^{2}75^{\circ}-\sin^{2}75^{\circ}=\cos 150^{\circ}=-\frac{\sqrt{3}}{2}$
g.$G=\frac{1-\tan 15^{\circ}}{1+\tan 15^{\circ}}=\frac{\tan 45^{\circ}-\tan 15^{\circ}}{\tan 45^{\circ}+\tan 15^{\circ}}=\tan\left( 45^{\circ}-15^{\circ} \right)=\tan 30^{\circ}=\frac{\sqrt{3}}{3}$
h.$H=\tan 15^{\circ}+\cot 15^{\circ}=\frac{\sin 15^{\circ}}{\cos 15^{\circ}}+\frac{\cos 15^{\circ}}{\sin 15^{\circ}}=\frac{\sin^{2}15^{\circ}+\cos^{2}15^{\circ}}{\sin 15^{\circ}.\cos 15^{\circ}}=\frac{1}{\frac{1}{2}\sin 2.15^{\circ}}=\frac{2}{\sin 30^{\circ}}=4$
$\begin{array}{l}A=\sin^{2}20^{\circ}+\sin^{2}100^{\circ}+\sin^{2}140^{\circ}\\=\sin^{2}20^{\circ}+\sin^{2}80^{\circ}+\sin^{2}40^{\circ}\\=\frac{1-\cos 40^{\circ}}{2}+\frac{1-\cos 160^{\circ}}{2}+\frac{1-\cos 80^{\circ}}{2}\\=\frac{3}{2}-\frac{\cos 40^{\circ}+\cos 160^{\circ}+\cos 80^{\circ}}{2}\end{array}$
Mà ta có:
$\begin{array}{l}\cos 40^{\circ}+\cos 160^{\circ}+\cos 80^{\circ}=\left( \cos 40^{\circ}+\cos 80^{\circ} \right)+\cos 160^{\circ}\\=2\cos 60^{\circ}\cos 20^{\circ}-\cos 20^{\circ}\\=\cos 20^{\circ}\left( 2\cos 60^{\circ}-1 \right)=0\end{array}$
Vậy $A=\frac{3}{2}$.
b.Tính $B=\cos^{2}10^{\circ}+\cos^{2}110^{\circ}+\cos^{2}130^{\circ}$
$\begin{array}{l}B=\cos^{2}10^{\circ}+\cos^{2}110^{\circ}+\cos^{2}130^{\circ}=\cos^{2}10^{\circ}+\cos^{2}70^{\circ}+\cos^{2}50^{\circ}\\=\frac{1+\cos 20^{\circ}}{2}+\frac{1+\cos 140^{\circ}}{2}+\frac{1+\cos 100^{\circ}}{2}\end{array}$
$\begin{array}{l}=\frac{3}{2}+\frac{\left( \cos 20^{\circ}+\cos 100^{\circ} \right)+\cos 140^{\circ}}{2}\\=\frac{3}{2}+\frac{2\cos 40^{\circ}.\cos 60^{\circ}-\cos 40^{\circ}}{2}\\=\frac{3}{2}+\frac{\cos 40^{\circ}\left( 2\cos 60^{\circ}-1 \right)}{2}=\frac{3}{2}\end{array}$
c.Tính $C=\tan 20^{\circ}.\tan 80^{\circ}+\tan 80^{\circ}.\tan 140^{\circ}+\tan 140^{\circ}.\tan 20^{\circ}$
Ta chứng minh công thức sau
$\tan x.\tan\left( x+\frac{\pi}{3} \right)+\tan\left( x+\frac{\pi}{3} \right).\tan\left( x+\frac{2\pi}{3} \right)+\tan x.\tan\left( x+\frac{2\pi}{3} \right)=-3$
Nhận Xét:$\tan\left( a-b \right)=\frac{\tan a-\tan b}{1+\tan a.\tan b}\Rightarrow \tan a.\tan b=\frac{\tan a-\tan b}{\tan\left( a-b \right)}-1$
Do vậy:
$\tan x.\tan\left( x+\frac{\pi}{3} \right)=\frac{\tan\left( x+\frac{\pi}{3} \right)-\tan x}{\tan\frac{\pi}{3}}-1=\frac{1}{\sqrt{3}}\left[ \tan\left( x+\frac{\pi}{3} \right)-\tan x \right]-1$ $\left( * \right)$
$\tan\left( x+\frac{\pi}{3} \right).\tan\left( x+\frac{2\pi}{3} \right)=\frac{\tan\left( x+\frac{2\pi}{3} \right)-\tan\left( x+\frac{\pi}{3} \right)}{\tan\frac{\pi}{3}}-1$
$=\frac{1}{\sqrt{3}}\left[ \tan\left( x+\frac{2\pi}{3} \right)-\tan\left( x+\frac{\pi}{3} \right) \right]-1$ $\left( ** \right)$
$\tan\left( x+\frac{2\pi}{3} \right).\tan\left( x \right)=\frac{\tan\left( x+\frac{2\pi}{3} \right)-\tan x}{\tan\frac{2\pi}{3}}-1=-\frac{1}{\sqrt{3}}\left[ \tan\left( x+\frac{2\pi}{3} \right)-\tan x \right]-1$ $\left( *** \right)$
Cộng theo vế $\left( * \right)$ $\left( ** \right)$ $\left( *** \right)$ ta được:
$\tan x.\tan\left( x+\frac{\pi}{3} \right)+\tan\left( x+\frac{\pi}{3} \right).\tan\left( x+\frac{2\pi}{3} \right)+\tan x.\tan\left( x+\frac{2\pi}{3} \right)=-3$
Vậy $C=\tan 20^{\circ}.\tan 80^{\circ}+\tan 80^{\circ}.\tan 140^{\circ}+\tan 140^{\circ}.\tan 20^{\circ}=-3$
d.Tương tự câu c
e.$E=\frac{\cot 225^{\circ}-\cot 79^{\circ}.\cot 71^{\circ}}{\cot 259^{\circ}+\cot 251^{\circ}}$
$\cot 225^{\circ}=\cot\left( 180^{\circ}+45^{\circ} \right)=\cot 45^{\circ}$
$\cot 79^{\circ}=\tan 11^{\circ}$
$\cot 71^{\circ}=\tan 19^{\circ}$
$\cot 259^{\circ}=\cot\left( 180^{\circ}+79^{\circ} \right)=\cot 79^{\circ}=\tan 11^{\circ}$
$\cot 251^{\circ}=\cot\left( 180^{\circ}+71^{\circ} \right)=\cot\left( 71^{\circ} \right)=\tan 19^{\circ}$
Vậy $E=\frac{\cot 225^{\circ}-\cot 79^{\circ}.\cot 71^{\circ}}{\cot 259^{\circ}+\cot 251^{\circ}}=\frac{1-\tan 11^{\circ}.\tan 19^{\circ}}{\tan 11^{\circ}+\tan 19^{\circ}}=\cot\left( 11^{\circ}+19^{\circ} \right)=\cot 30^{\circ}=\sqrt{3}$
f.$F=\cos^{2}75^{\circ}-\sin^{2}75^{\circ}=\cos 150^{\circ}=-\frac{\sqrt{3}}{2}$
g.$G=\frac{1-\tan 15^{\circ}}{1+\tan 15^{\circ}}=\frac{\tan 45^{\circ}-\tan 15^{\circ}}{\tan 45^{\circ}+\tan 15^{\circ}}=\tan\left( 45^{\circ}-15^{\circ} \right)=\tan 30^{\circ}=\frac{\sqrt{3}}{3}$
h.$H=\tan 15^{\circ}+\cot 15^{\circ}=\frac{\sin 15^{\circ}}{\cos 15^{\circ}}+\frac{\cos 15^{\circ}}{\sin 15^{\circ}}=\frac{\sin^{2}15^{\circ}+\cos^{2}15^{\circ}}{\sin 15^{\circ}.\cos 15^{\circ}}=\frac{1}{\frac{1}{2}\sin 2.15^{\circ}}=\frac{2}{\sin 30^{\circ}}=4$