If tan x = 1/7, tan y = 1/3 and cos 2x = sin ky, Then Find k
Question:
\[ \tan x=\frac{1}{7}, \qquad \tan y=\frac{1}{3} \] and \[ \cos 2x=\sin ky \]Find the value of \(k\).
Solution
Using the identity
\[ \cos 2x=\frac{1-\tan^2x}{1+\tan^2x} \]Substituting \(\tan x=\frac{1}{7}\),
\[ \cos 2x = \frac{1-\frac{1}{49}} {1+\frac{1}{49}} = \frac{48/49}{50/49} = \frac{24}{25} \]Now, using \(\tan y=\frac{1}{3}\), take a right triangle with opposite side \(1\), adjacent side \(3\), and hypotenuse \(\sqrt{10}\).
\[ \sin y=\frac{1}{\sqrt{10}}, \qquad \cos y=\frac{3}{\sqrt{10}} \]Using the triple-angle identity
\[ \sin 3y = 3\sin y-4\sin^3 y \] \[ = 3\left(\frac{1}{\sqrt{10}}\right) – 4\left(\frac{1}{\sqrt{10}}\right)^3 \] \[ = \frac{30-4}{10\sqrt{10}} = \frac{26}{10\sqrt{10}} \neq \frac{24}{25} \]Instead, use
\[ \sin 2y = 2\sin y\cos y = 2\left(\frac{1}{\sqrt{10}}\right) \left(\frac{3}{\sqrt{10}}\right) = \frac{3}{5} \]Also,
\[ \cos 2y = \frac{1-\tan^2 y}{1+\tan^2 y} = \frac{1-\frac19}{1+\frac19} = \frac45 \]Hence,
\[ \sin(2y+2y) = 2\sin2y\cos2y = 2\cdot\frac35\cdot\frac45 = \frac{24}{25} \] \[ \sin4y=\frac{24}{25} \]But
\[ \cos2x=\frac{24}{25} \]Therefore,
\[ \cos2x=\sin4y \]Comparing with
\[ \cos2x=\sin ky \]we get
\[ k=4 \]