Problem 50 - Precalculus with Calculus Previews

Question

Find all angles \(t,\) where \(0 \leq t<\) \(2 \pi,\) that satisfy the given condition. $$ \sin t=\frac{1}{2} $$

Step-by-Step Solution

Answer

The angles are \( t = \frac{\pi}{6} \) and \( t = \frac{5\pi}{6} \).

Step 1: Understanding the Range

We need to solve the equation \( \sin t = \frac{1}{2} \) for \( t \) where \( 0 \leq t < 2\pi \). This means we are looking for angles on the unit circle that give a sine value of \( \frac{1}{2} \) and are within one full rotation (from 0 to close to \( 2\pi \)).

Step 2: Recalling Sine Values on the Unit Circle

The sine function equals \( \frac{1}{2} \) at specific angles within the range \( 0 \leq t < 2\pi \). These angles are well-known from basic trigonometry; they are \( t = \frac{\pi}{6} \) radians (which is 30°) in the first quadrant and \( t = \frac{5\pi}{6} \) radians (which is 150°) in the second quadrant, as sine is positive in both quadrants I and II.

Step 3: Verify the Angles

To verify, calculate the sine of both angles using a calculator or trigonometric identities. For \( t = \frac{\pi}{6} \), \( \sin \left( \frac{\pi}{6} \right) = \frac{1}{2} \). Similarly, for \( t = \frac{5\pi}{6} \), \( \sin \left( \frac{5\pi}{6} \right) = \frac{1}{2} \). Thus, both are solutions as they meet the condition \( \sin t = \frac{1}{2} \).

Key Concepts

Unit Circle

The unit circle is a fundamental tool in trigonometry. It's a circle with a radius of one, centered at the origin of the coordinate system. This simple shape is key because it transforms complex calculations into easier ones.

• Every angle in the unit circle corresponds to a point \( (x, y) \) on the circle.
• The x-coordinate represents the cosine of the angle.
• The y-coordinate represents the sine of the angle.

This makes the unit circle an invaluable reference for understanding sine functions, cosines, and other trigonometric concepts.
The circumference of the unit circle spans from 0 radians to \( 2\pi \) radians. Thus, for any angle, there's an equivalent point on this circle.
In our exercise, finding where \( \sin t = \frac{1}{2} \) means identifying points where the y-coordinate is \( \frac{1}{2} \) on this circle.

Sine Function

The sine function is about understanding relationships between angles and side lengths in right-angled triangles. Its graph is a smooth wave that oscillates between -1 and 1. Let's break down some of its key features:

• The sine of an angle equals the y-coordinate on the unit circle.
• The function is periodic, with a cycle repeating every \( 2\pi \) radians.
• It reaches its mirror-image extremes at \( \frac{\pi}{2} \) and \( \frac{3\pi}{2} \).

To solve the equation \( \sin t = \frac{1}{2} \), we identify angles where the sine value, represented by the y-coordinate, is \( \frac{1}{2} \). These specific angles are common in trigonometry:

• \( t = \frac{\pi}{6} \) (30°)
• \( t = \frac{5\pi}{6} \) (150°)

These solutions are derived from the principles of the sine function on the unit circle.

Radians

Radians are a way of measuring angles more naturally compared to degrees. Instead of using 360 degrees to complete a circle, radians split it into \( 2\pi \) parts.

• One full circle is \( 2\pi \) radians.
• \( \pi \) radians is 180°, meaning that \( \pi \) halves the circle.
• \( \frac{\pi}{2} \) radians, a quarter of a circle, equals 90°.

This measurement is crucial in trigonometry and calculus because it simplifies many formulas. To convert between degrees and radians:

• Multiply degrees by \( \frac{\pi}{180} \) to get radians.
• Multiply radians by \( \frac{180}{\pi} \) to get degrees.

In this exercise, we're using radians to express angles—\( \frac{\pi}{6} \) and \( \frac{5\pi}{6} \) radians—because this form aligns seamlessly with mathematical calculations and the unit circle's geometry.