Bài tập nâng cao · Bài 4
Dạng 1. Công thức cộng
Bài tập nâng cao Công thức lượng giác · Bài 4
Chứng minh các hệ thức sau với điều kiện cho trước
a.)$2\tan a=\tan\left( a+b \right)$ khi $\sin b=\sin a.\cos\left( a+b \right)$
b.)$2\tan a=\tan\left( a+b \right)$ khi $3\sin b=\sin\left( 2a+b \right)$
c.)$\tan a.\tan b=-\frac{1}{3}$ khi $\cos\left( a+b \right)=2\cos\left( a-b \right)$
d.)$\tan\left( a+b \right).\tan b=\frac{1-k}{1+k}$ khi $\cos\left( a+2b \right)=k.\cos a$
a.)$2\tan a=\tan\left( a+b \right)$ khi $\sin b=\sin a.\cos\left( a+b \right)$
b.)$2\tan a=\tan\left( a+b \right)$ khi $3\sin b=\sin\left( 2a+b \right)$
c.)$\tan a.\tan b=-\frac{1}{3}$ khi $\cos\left( a+b \right)=2\cos\left( a-b \right)$
d.)$\tan\left( a+b \right).\tan b=\frac{1-k}{1+k}$ khi $\cos\left( a+2b \right)=k.\cos a$
Xem lời giải
Lời giải
a. Ta có
$\begin{array}{l}2\tan a=\tan\left( a+b \right)\Leftrightarrow \frac{2\sin a}{\cos a}=\frac{\sin\left( a+b \right)}{\cos\left( a+b \right)}\\\Leftrightarrow \frac{2\sin a}{\cos a}=\frac{\sin\left( a+b \right)}{\cos\left( a+b \right)}\\\Leftrightarrow \frac{2\sin a\cos\left( a+b \right)-\cos a.\sin\left( a+b \right)}{\cos a\cos\left( a+b \right)}=0\\\Leftrightarrow \frac{\sin a\cos\left( a+b \right)+\sin a\cos\left( a+b \right)-\cos a.\sin\left( a+b \right)}{\cos a\cos\left( a+b \right)}=0\\\Leftrightarrow \frac{\sin a\cos\left( a+b \right)+\sin\left( a-a-b \right)}{\cos a\cos\left( a+b \right)}=0\Leftrightarrow \frac{\sin b-\sin b}{\cos a\cos\left( a+b \right)}=0\end{array}$
Vậy ta có điều phải chứng minh
b. Ta có:
$\begin{array}{l}2\tan a-\tan\left( a+b \right)=\frac{2\sin a}{\cos a}-\frac{\sin\left( a+b \right)}{\cos\left( a+b \right)}\\=\frac{2\sin a}{\cos a}-\frac{\sin a\cos b+\sin b\cos a}{\cos a\cos b-\sin a\sin b}\\=\frac{2\sin a\cos a\cos b-2\sin^{2}a\sin b-\sin a\cos a\cos b-\sin b\cos^{2}a}{\cos a.\left( \cos a\cos b-\sin a\sin b \right)}\end{array}$
$\begin{array}{l}=\frac{\sin a\cos a\cos b-\sin b-\sin^{2}a\sin b}{\cos a.\left( \cos a\cos b-\sin a\sin b \right)}\\=\frac{-\sin b+\sin a\cos a\cos b-\sin^{2}a\sin b}{\cos a.\left( \cos a\cos b-\sin a\sin b \right)}\end{array}$
$\begin{array}{l}=\frac{-2\sin b+2\sin a\cos a\cos b-2\sin^{2}a\sin b}{2\cos a.\left( \cos a\cos b-\sin a\sin b \right)}\\=\frac{-3\sin b+2\sin a\cos a\cos b+\sin b-2\sin^{2}a\sin b}{2\cos a.\left( \cos a\cos b-\sin a\sin b \right)}=\frac{-3\sin b+\sin 2a\cos b+\sin b\left( 1-2\sin^{2}a \right)}{2\cos a.\left( \cos a\cos b-\sin a\sin b \right)}\\=\frac{-3\sin b+\sin 2a\cos b+\sin b\cos 2a}{2\cos a.\left( \cos a\cos b-\sin a\sin b \right)}=\frac{-3\sin b+\sin\left( 2a+b \right)}{2\cos a.\left( \cos a\cos b-\sin a\sin b \right)}=0\end{array}$
Vậy đẳng thức được chứng minh.
c. Ta có: $\tan a\tan b=\frac{\sin a\sin b}{\cos a\cos b}=\frac{\cos\left( a-b \right)-2\cos\left( a-b \right)}{\cos\left( a-b \right)+2\cos\left( a-b \right)}=\frac{-\cos\left( a-b \right)}{3\cos\left( a-b \right)}=-\frac{1}{3}$
d. Ta có: $\tan\left( a+b \right)\tan b=\frac{\sin\left( a+b \right)\sin b}{\cos\left( a+b \right)\cos b}$ $=\frac{\cos a-\cos\left( a+2b \right)}{\cos a+\cos\left( a+2b \right)}=\frac{\cos a-k\cos a}{\cos a+k\cos a}=\frac{\left( 1-k \right)\cos a}{\left( 1+k \right)\cos a}=\frac{1-k}{1+k}$
$\begin{array}{l}2\tan a=\tan\left( a+b \right)\Leftrightarrow \frac{2\sin a}{\cos a}=\frac{\sin\left( a+b \right)}{\cos\left( a+b \right)}\\\Leftrightarrow \frac{2\sin a}{\cos a}=\frac{\sin\left( a+b \right)}{\cos\left( a+b \right)}\\\Leftrightarrow \frac{2\sin a\cos\left( a+b \right)-\cos a.\sin\left( a+b \right)}{\cos a\cos\left( a+b \right)}=0\\\Leftrightarrow \frac{\sin a\cos\left( a+b \right)+\sin a\cos\left( a+b \right)-\cos a.\sin\left( a+b \right)}{\cos a\cos\left( a+b \right)}=0\\\Leftrightarrow \frac{\sin a\cos\left( a+b \right)+\sin\left( a-a-b \right)}{\cos a\cos\left( a+b \right)}=0\Leftrightarrow \frac{\sin b-\sin b}{\cos a\cos\left( a+b \right)}=0\end{array}$
Vậy ta có điều phải chứng minh
b. Ta có:
$\begin{array}{l}2\tan a-\tan\left( a+b \right)=\frac{2\sin a}{\cos a}-\frac{\sin\left( a+b \right)}{\cos\left( a+b \right)}\\=\frac{2\sin a}{\cos a}-\frac{\sin a\cos b+\sin b\cos a}{\cos a\cos b-\sin a\sin b}\\=\frac{2\sin a\cos a\cos b-2\sin^{2}a\sin b-\sin a\cos a\cos b-\sin b\cos^{2}a}{\cos a.\left( \cos a\cos b-\sin a\sin b \right)}\end{array}$
$\begin{array}{l}=\frac{\sin a\cos a\cos b-\sin b-\sin^{2}a\sin b}{\cos a.\left( \cos a\cos b-\sin a\sin b \right)}\\=\frac{-\sin b+\sin a\cos a\cos b-\sin^{2}a\sin b}{\cos a.\left( \cos a\cos b-\sin a\sin b \right)}\end{array}$
$\begin{array}{l}=\frac{-2\sin b+2\sin a\cos a\cos b-2\sin^{2}a\sin b}{2\cos a.\left( \cos a\cos b-\sin a\sin b \right)}\\=\frac{-3\sin b+2\sin a\cos a\cos b+\sin b-2\sin^{2}a\sin b}{2\cos a.\left( \cos a\cos b-\sin a\sin b \right)}=\frac{-3\sin b+\sin 2a\cos b+\sin b\left( 1-2\sin^{2}a \right)}{2\cos a.\left( \cos a\cos b-\sin a\sin b \right)}\\=\frac{-3\sin b+\sin 2a\cos b+\sin b\cos 2a}{2\cos a.\left( \cos a\cos b-\sin a\sin b \right)}=\frac{-3\sin b+\sin\left( 2a+b \right)}{2\cos a.\left( \cos a\cos b-\sin a\sin b \right)}=0\end{array}$
Vậy đẳng thức được chứng minh.
c. Ta có: $\tan a\tan b=\frac{\sin a\sin b}{\cos a\cos b}=\frac{\cos\left( a-b \right)-2\cos\left( a-b \right)}{\cos\left( a-b \right)+2\cos\left( a-b \right)}=\frac{-\cos\left( a-b \right)}{3\cos\left( a-b \right)}=-\frac{1}{3}$
d. Ta có: $\tan\left( a+b \right)\tan b=\frac{\sin\left( a+b \right)\sin b}{\cos\left( a+b \right)\cos b}$ $=\frac{\cos a-\cos\left( a+2b \right)}{\cos a+\cos\left( a+2b \right)}=\frac{\cos a-k\cos a}{\cos a+k\cos a}=\frac{\left( 1-k \right)\cos a}{\left( 1+k \right)\cos a}=\frac{1-k}{1+k}$