Throughout this chapter it will be assumed that the student is familiar with the unit circle and the values of the sin𝜃,cos𝜃, and tan𝜃 for the common angles measurements in both radians and degrees such as $0°$, $30°\left(\dfrac{\pi}{6}\right)$, $45°\left(\dfrac{\pi}{4}\right)$, $60°\left(\dfrac{\pi}{3}\right)$, $\ldots$ It is also assumed that the students understands whether $\sin\theta$, $\cos\theta$, and $\tan\theta$ are positive or negative in all four quadrants. This knowledge will be imperative in answering these questions involving equations and double angles.

In any triangle let $\theta$ be an angle. Without going into a formal proof, it is known that:

$\tan\theta = \dfrac{\sin\theta}{\cos\theta}$,
$\sin^2\theta + \cos^2\theta = 1$,
$\sin(2\theta) = 2\sin\theta\cos\theta$,
$\cos(2\theta)=\cos^2\theta − \sin^2\theta$,
$\cos(2\theta)=2 \cos^2\theta − 1$,
$\cos(2\theta) = 1 − 2 \sin^2\theta$.

Recall that in quadrant I, all three trig functions are positive. In quadrant II only the sin function is positive. In quadrant III only the tan function is positive, and in quadrant IV, only the cos function is positive.

Example 1: If $\tan\theta=−\dfrac{7}{24}$, and $\dfrac{\pi}{2} < \theta < \pi$, determine the value of $\cos(2\theta)$

Explanation: It will be helpful to sketch a right triangle in quad II where the opposite leg has length $7$ and the adjacent leg has length $24$. From there, using the Pythagorean Theorem, we can deduce that the hypotenuse has length $25$. This implies that $\sin\theta=\dfrac{7}{25}$ and $\cos\theta =−\dfrac{24}{25}$ (because in quad II, cos is negative). Thus, using $1$ of the $3$ formulas for $\cos(2\theta)$ gives:

$\cos(2\theta)=2\cos^2\theta − 1$ $= 2\left(−\dfrac{24}{25}\right)^2−1=\dfrac{527}{625}$

Example 2: Solve for $x$ in $\sqrt 2\sin x=1$ on a domain of $0< x < 2\pi$

Explanation: Before solving for $x$, it is often a good idea to solve for the given trig function first, then recall the unit circle measurements of this trig function. Thus:

$\sqrt 2\sin x = 1$
$\sin x = \dfrac{1}{\sqrt 2}$
$\sin^{−1}\left(\dfrac{1}{\sqrt 2}\right)=45°$ 

We can see this using our TI-84 CE graphing calculator:

However, this is not the complete solution. Remember that the sine of an angle is positive in quadrants I and II. The calculator only gives one answer. It will only give us the answer in quadrant I. The sine of an angle is also $\sqrt{1}{\sqrt 2}$ at $135°$. Therefore the correct solution is $x=45°$ and $x=135°$.