Problem 74 - Precalculus Essentials

Question

In Exercises \(61-86,\) use reference angles to find the exact value of each expression. Do not use a calculator. $$\sin \left(-225^{\circ}\right)$$

Step-by-Step Solution

Answer

The exact value of \(\sin \left(-225^{\circ}\right)\) is \(\frac{\sqrt{2}}{2}\)

Step 1: Determine the position of the angle

The given angle is \(-225^{\circ}\). Trigonometric functions have a period of \(360^{\circ}\) or \(2\pi \) radian, so they repeat the same set of values for every full rotation. For an angle greater than \(180^{\circ}\), you can subtract \(360^{\circ}\) until you get a negative acute angle in the fourth quadrant. So, \(-225^{\circ} + 360^{\circ} = 135^{\circ}\), therefore \(-225^{\circ} \) and \(135^{\circ}\) are coterminal, meaning that they are in the same place on the unit circle.

Step 2: Determine the reference angle

A reference angle for any angle is the acute angle to the x-axis. Even though \(-225^{\circ} \) and \(135^{\circ}\) both land on the same point in the unit circle, their reference angles are still \(45^{\circ}\) in either case. Note that for any angle, the reference angle is always positive.

Step 3: Determine the sign of the trigonometric function

The reference angle sits in the second quadrant. As you may recall, sine is positive in the second quadrant.

Step 4: Find the exact value using the reference angle

The reference angle already places the angle in the context of a right triangle, and now you simply find the sine of \(45^{\circ}\). That's an angle we know the sine of: \(\sin(45^{\circ}) = \frac{1}{\sqrt{2}}\) or \(\sin(45^{\circ}) = \frac{\sqrt{2}}{2}\), if you rationalize the denominator.