Problem 111 - Precalculus

Question

For the following exercises, find the exact value using half-angle formulas. $$ \sin \left(\frac{\pi}{8}\right) $$

Step-by-Step Solution

Answer

\(\sin\left(\frac{\pi}{8}\right) = \frac{\sqrt{2 - \sqrt{2}}}{2}\)

Step 1: Understand the Half-Angle Formula

The half-angle formula for the sine function is given by \( \sin\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos(\theta)}{2}} \). We need to use this formula to find \( \sin\left(\frac{\pi}{8}\right) \), noting that \( \frac{\pi}{8} = \frac{\pi/4}{2} \). Hence, \( \theta = \frac{\pi}{4} \).

Step 2: Find \(\cos(\theta)\)

We need \( \cos\left(\frac{\pi}{4}\right) \) to use in our half-angle formula. The value of \( \cos\left(\frac{\pi}{4}\right) \) is \( \frac{\sqrt{2}}{2} \).

Step 3: Substitute Values and Simplify

Plug \( \theta = \frac{\pi}{4} \) into the half-angle formula: \( \sin\left(\frac{\pi}{8}\right) = \pm \sqrt{\frac{1 - \cos\left(\frac{\pi}{4}\right)}{2}} \). Substitute \( \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \) to get \( \sin\left(\frac{\pi}{8}\right) = \pm \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}} \).

Step 4: Simplify the Expression Further

Calculate \( 1 - \frac{\sqrt{2}}{2} = \frac{2 - \sqrt{2}}{2} \). Substitute back to get \( \sin\left(\frac{\pi}{8}\right) = \pm \sqrt{\frac{2 - \sqrt{2}}{4}} \). Simplify to \( \sin\left(\frac{\pi}{8}\right) = \pm \frac{\sqrt{2 - \sqrt{2}}}{2} \).

Step 5: Determine the Sign

Since \( \frac{\pi}{8} \) is in the first quadrant, where the sine is positive, we choose the positive value. Therefore, \( \sin\left(\frac{\pi}{8}\right) = \frac{\sqrt{2 - \sqrt{2}}}{2} \).

Key Concepts

Trigonometric Identities

Trigonometric identities are essential mathematical equations involving trigonometric functions like sine, cosine, and tangent. These equations hold true for all values of the variables involved and express relationships between the different trigonometric functions. A key identity is the Pythagorean identity, which states that \( \sin^2(x) + \cos^2(x) = 1 \). This identity forms the basis for many other trigonometric equations.
One of the most useful groups of trigonometric identities for solving problems such as the one given is the family of half-angle identities. These allow you to find the value of a trigonometric function at half of a given angle. The half-angle formula for sine, which is used in the original problem, is:
\[ \sin\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos(\theta)}{2}} \]
These formulas are very helpful because they can convert complex trigonometric functions into forms that are easier to evaluate.

Exact Values in Trigonometry

Exact values in trigonometry refer to the specific, non-decimal values of trigonometric functions for certain angles. These can be found using key angles such as \(0^{\circ}, 30^{\circ}, 45^{\circ}, 60^{\circ},\) and \(90^{\circ}\), and their radian equivalents. For instance, \( \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \) is an exact value.
When using the half-angle identity, obtaining the exact value requires substituting known exact values from the basic angle formulas. As seen in the problem, knowing \( \cos(\frac{\pi}{4}) \) is crucial to finding \( \sin(\frac{\pi}{8}) \).
These exact values are derived from the geometry of triangles and the unit circle, which allows them to be used confidently without approximation. Understanding them helps simplify expressions during problem-solving and provides precise solutions without the need for a calculator.

Trigonometric Functions

Trigonometric functions are foundational to understanding the angles and sides of triangles, as well as periodic phenomena. The primary functions are sine (\sin), cosine (\cos), and tangent (\tan). Each function relates an angle in a right-angled triangle to the ratio of two of its sides.
The sine function specifically measures the ratio of the length of the side opposite the angle to the hypotenuse. In the original exercise, we calculated \(\sin\left(\frac{\pi}{8}\right)\), which shows how these functions extend beyond the immediate context of triangles to more complex mathematical analysis.
These functions are pivotal in representing oscillatory or wave-like phenomena in physics, engineering, and various scientific fields. They also form the basis for more advanced functions and equations in both trigonometry and calculus, making a good grasp of them crucial for advanced mathematical studies.