Chapter 12.3, Problem 35WE

To determine

Write the given as a function of an acute angle.

Answer to Problem 35WE

  $-\sin 32.67°$

Explanation of Solution

Given:

  $\sin 212°40\text{'}$

Formula Used:

If $\theta$ is not a quadrantal angle, there is a unique acute angle $\alpha$ corresponding to $\theta$ , formed by the terminal side of $\theta$ and the positive or negative x- axis.When $\theta$ is in standard position , $\alpha$ is called the reference angle of $\theta$ .

Function value	Quadrant of $\theta$ Is I	Quadrant of $\theta$ Is II	Quadrant of $\theta$ Is III	Quadrant of $\theta$ Is IV
$\begin{array}{l}\sin \theta \\ \csc \theta \end{array}$	+	+	-	-
$\begin{array}{l}\cos \theta \\ \sec \theta \end{array}$	+	-	-	+
$\begin{array}{l}\tan \theta \\ \cot \theta \end{array}$	+	-	+	-

Calculation:

Since, $\theta =212°40\text{'}$ ,first convert minutes into degree:

  $\theta =212°40\text{'}=212°+\left(\frac{40}{60}\right)°=212°+0.67°=212.67°$ , which is in Quadrant III, where sin is negative.

So, the reference angle is :

  $\alpha =\theta \text{'}-180°=212.67°-180°=32.67°$

So, the given angle as a function of acute angle is :

  $-\sin 32.67°$