Problem 19 - Precalculus Mathematics for Calculus

Question

Use an appropriate Half-Angle Formula to find the exact value of the expression. $$\tan 22.5^{\circ}$$

Step-by-Step Solution

Answer

\( \tan 22.5^{\circ} = \sqrt{3 - 2\sqrt{2}} \).

Step 1: Recall the Half-Angle Formula for Tangent

The half-angle formula for tangent is given by \( \tan \frac{θ}{2} = \pm \sqrt{\frac{1 - \cos θ}{1 + \cos θ}} \). Since we need to find \( \tan 22.5^{\circ} \), we can express this angle as \( 45^{\circ}/2 \). This allows us to use the half-angle formula for \( \tan \frac{45^{\circ}}{2} \).

Step 2: Insert Known Values into the Formula

Using the half-angle formula \( \tan \frac{θ}{2} = \pm \sqrt{\frac{1 - \cos θ}{1 + \cos θ}} \), we substitute \( θ = 45^{\circ} \). The cosine of 45 degrees is known: \( \cos 45^{\circ} = \frac{\sqrt{2}}{2} \).

Step 3: Solve for \( \tan 22.5^{\circ} \)

Now, substitute \( \cos 45^{\circ} = \frac{\sqrt{2}}{2} \) into the formula: \( \tan 22.5^{\circ} = \pm \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}}} \).

Step 4: Simplify the Expression

Simplify the numerator: \( 1 - \frac{\sqrt{2}}{2} = \frac{2}{2} - \frac{\sqrt{2}}{2} = \frac{2-\sqrt{2}}{2} \). Similarly, simplify the denominator: \( 1 + \frac{\sqrt{2}}{2} = \frac{2}{2} + \frac{\sqrt{2}}{2} = \frac{2+\sqrt{2}}{2} \).

Step 5: Further Simplify and Find Exact Value

The expression now becomes \( \tan 22.5^{\circ} = \sqrt{\frac{\frac{2-\sqrt{2}}{2}}{\frac{2+\sqrt{2}}{2}}} = \sqrt{\frac{2-\sqrt{2}}{2+\sqrt{2}}} \). To rationalize the denominator, multiply by the conjugate: \( \frac{2-\sqrt{2}}{2+\sqrt{2}} \times \frac{2-\sqrt{2}}{2-\sqrt{2}} \). This simplifies to \( \frac{(2-\sqrt{2})^2}{(2+\sqrt{2})(2-\sqrt{2})} \).

Step 6: Final Calculation

Calculate the final value: \( (2-\sqrt{2})^2 = 4 - 4\sqrt{2} + 2 = 6 - 4\sqrt{2} \) and \( (2+\sqrt{2})(2-\sqrt{2}) = 4 - 2 = 2 \). Thus, \( \tan 22.5^{\circ} = \sqrt{\frac{6-4\sqrt{2}}{2}} = \sqrt{3 - 2\sqrt{2}} \).

Step 7: Confirm the Sign

At \( 22.5^{\circ} \), tangent is positive because it is in the first quadrant. Therefore, the solution is \( \tan 22.5^{\circ} = \sqrt{3 - 2\sqrt{2}} \).

Key Concepts

Tangent Function

The tangent function is a fundamental part of trigonometry, often associated with right-angled triangles and the unit circle. The tangent of an angle in a triangle is defined as the ratio of the length of the opposite side to the length of the adjacent side. Mathematically, for an angle \( \theta \), this is expressed as:\[\tan \theta = \frac{\text{opposite}}{\text{adjacent}}\]When dealing with the unit circle, which is a circle with a radius of one, the tangent can be represented as the sine of the angle divided by the cosine of the angle:\[\tan \theta = \frac{\sin \theta}{\cos \theta}\]Understanding this relationship helps in exploring more complex trigonometric identities and formulas, such as the half-angle formulas, which are particularly useful in finding angles like \( \tan 22.5^{\circ} \). This specific formula is:\[\tan \frac{\theta}{2} = \pm"][\sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}}\]The sign depends on the quadrant where the angle \( \theta \) lies. Using the half-angle formula allows simplification by translating complex angle calculations into simpler ones.

Cosine Function

The cosine function is another fundamental trigonometric function that relates the angle of a right-angled triangle to the lengths of the adjacent side and the hypotenuse. For an angle \( \theta \), cosine is given by:\[\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}\]On the unit circle, the cosine of an angle \( \theta \) is represented as the x-coordinate of the corresponding point on the circle. This interpretation is pivotal when considering angles beyond standard right triangles, and it connects directly with periodic functions and waves.For half-angle calculations, knowing \( \cos 45^{\circ} = \frac{\sqrt{2}}{2} \) helped derive the value of \( \tan 22.5^{\circ} \). Understanding and memorizing key values such as \( \cos 30^{\circ}, \cos 45^{\circ}, \) and \( \cos 60^{\circ} \) are essential in solving trigonometric equations using identities:

• \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \)
• \( \cos 45^{\circ} = \frac{\sqrt{2}}{2} \)
• \( \cos 60^{\circ} = \frac{1}{2} \)

In the specific problem of finding \( \tan 22.5^{\circ} \), utilizing these cosine values within the half-angle formula simplified the process.

Trigonometric Identities

Trigonometric identities are mathematical equations that relate the trigonometric functions of an angle. These identities are pivotal not just in solving trigonometric equations but also in simplifying expressions involving angles. Some of the core identities include:

• Pythagorean Identity: \( \sin^2 \theta + \cos^2 \theta = 1 \)
• Quotient Identity: \( \tan \theta = \frac{\sin \theta}{\cos \theta} \)
• Reciprocal Identities: \( \csc \theta = \frac{1}{\sin \theta}, \sec \theta = \frac{1}{\cos \theta}, \cot \theta = \frac{1}{\tan \theta} \)

Half-angle identities, like the one used here for tangent, allow for simplifying the calculation of angles that are half of common benchmark angles, such as \( 22.5^{\circ} \), which is half of \( 45^{\circ} \). These identities become significantly useful when exact values are required, expanding our toolkit beyond numerical trigonometric tables.Mastering these identities and understanding their derivations strengthen problem-solving skills in trigonometry. It enables a deeper understanding of how angles interact in more complex mathematical and real-world scenarios.