Chapter 8: The unit circle and radian measure

1. Suppose $\tan x = k$ where $k$ is a constant and $\pi<x<\frac{3\pi}{2}$ . Write expression for $\sin x$ and $\cos x$ in terms of $k.$

Solution:

$\because\:\:\:\:\cos^2 x + \sin^2 x = 1$

$\therefore\:\:\:\:\cos^2 x = 1 - \sin^2 x$

$\because\:\:\:\:\frac{\sin x}{\cos x} = \tan x$

$\therefore\:\:\:\:(\frac{\sin x}{\cos x})^2 = \tan^2 x$

$\therefore\:\:\:\:\frac{\sin^2 x}{\cos^2 x} = \tan^2 x$

$\therefore\:\:\:\:\frac{\sin^2 x}{1 - \sin^2 x} = k^2$

$\therefore\:\:\:\:\sin^2 x = (1 - \sin^2 x)k^2$

$\therefore\:\:\:\:\sin^2 x = k^2 - k^2\sin^2 x$

$\therefore\:\:\:\:\sin^2 x + k^2\sin^2 x = k^2$

$\therefore\:\:\:\:\sin^2 x(1 + k^2)= k^2$

$\therefore\:\:\:\:\sin^2 x= \frac{k^2}{1+ k^2}$

$\because\:\:\:\:\pi<x<\frac{3\pi}{2}$

$\therefore\:\:\:\:\sin x < 0$

$\therefore\:\:\:\:\sin x = -\sqrt{\frac{k^2}{1+ k^2}}$

$\because\:\:\:\:\cos^2 x = 1 - \sin^2 x$

$\therefore\:\:\:\:\cos^2 x = 1 - \frac{k^2}{1+ k^2}$

$\therefore\:\:\:\:\cos^2 x = \frac{1+k^2-k^2}{1+ k^2}$

$\therefore\:\:\:\:\cos^2 x = \frac{1}{1+ k^2}$

$\because\:\:\:\:\pi<x<\frac{3\pi}{2}$

$\therefore\:\:\:\:\cos x < 0$

$\therefore\:\:\:\:\cos x = -\sqrt{\frac{1}{1+ k^2}}$