If tan A = 1/7 and tan B = 1/3, Show that cos 2A = sin 4B
Question
If \[ \tan A=\frac{1}{7} \quad \text{and} \quad \tan B=\frac{1}{3}, \] show that \[ \cos 2A=\sin 4B. \]
Solution
Using the identity,
\[ \cos 2A=\frac{1-\tan^2A}{1+\tan^2A} \]
Substituting \[ \tan A=\frac{1}{7}, \] we get
\[ \cos 2A = \frac{1-\left(\frac{1}{7}\right)^2} {1+\left(\frac{1}{7}\right)^2} \]
\[ = \frac{1-\frac{1}{49}} {1+\frac{1}{49}} \]
\[ = \frac{\frac{48}{49}} {\frac{50}{49}} \]
\[ = \frac{48}{50} = \frac{24}{25} \]
Now, \[ \tan B=\frac{1}{3} \]
Using the identity,
\[ \sin 2B= \frac{2\tan B}{1+\tan^2B} \]
\[ = \frac{2\times\frac{1}{3}} {1+\left(\frac{1}{3}\right)^2} \]
\[ = \frac{\frac{2}{3}} {1+\frac{1}{9}} \]
\[ = \frac{\frac{2}{3}} {\frac{10}{9}} \]
\[ = \frac{3}{5} \]
Also,
\[ \cos 2B= \frac{1-\tan^2B}{1+\tan^2B} \]
\[ = \frac{1-\frac{1}{9}} {1+\frac{1}{9}} \]
\[ = \frac{\frac{8}{9}} {\frac{10}{9}} \]
\[ = \frac{4}{5} \]
Using \[ \sin 4B = 2\sin 2B \cos 2B \]
\[ = 2\times\frac{3}{5}\times\frac{4}{5} \]
\[ = \frac{24}{25} \]
Thus,
\[ \sin 4B=\frac{24}{25} \]
and
\[ \cos 2A=\frac{24}{25} \]
Therefore,
\[ \boxed{\cos 2A=\sin 4B} \]