Problem 21 - Algebra 2 and Trigonometry

Question

Use \(\cos A=\cos 30^{\circ}=\frac{\sqrt{3}}{2}\) to show that the exact value of \(\tan 15^{\circ}=2-\sqrt{3}\)

Step-by-Step Solution

Answer

\(\tan 15^\circ = 2 - \sqrt{3}\) using the half-angle formula.

Step 1: Recall the Half-Angle Identity for Tangent

The half-angle identity for tangent is given by: \[\tan\left(\frac{A}{2}\right) = \frac{1 - \cos A}{\sin A}\]Since we are given that \(A = 30^\circ\), we need to find \(\tan 15^\circ\), which is half of \(30^\circ\).

Step 2: Find Sine of the Given Angle

Use the Pythagorean identity to find \(\sin A\):\[\sin^2 A + \cos^2 A = 1\]Substitute \(\cos 30^\circ = \frac{\sqrt{3}}{2}\):\[\sin^2 30^\circ + \left(\frac{\sqrt{3}}{2}\right)^2 = 1\]Calculate \(\sin^2 30^\circ\):\[\sin^2 30^\circ + \frac{3}{4} = 1 \quad \Rightarrow \quad \sin^2 30^\circ = \frac{1}{4}\]Thus, \(\sin 30^\circ = \frac{1}{2}\).

Step 3: Substitute Values into the Half-Angle Formula

Using the half-angle formula:\[\tan 15^\circ = \frac{1 - \cos 30^\circ}{\sin 30^\circ}\]Substitute the known values:\[\tan 15^\circ = \frac{1 - \frac{\sqrt{3}}{2}}{\frac{1}{2}}\]

Step 4: Simplify the Expression

Simplify the fraction:\[\tan 15^\circ = \frac{2 \times \left(1 - \frac{\sqrt{3}}{2}\right)}{1}\]This simplifies to:\[\tan 15^\circ = 2 - \sqrt{3}\]

Step 5: Verify the Result

Recheck the calculation:Start with:\[\tan 15^\circ = \frac{2 \times 1 - 2 \times \frac{\sqrt{3}}{2}}{1} = 2 - \sqrt{3}\]Everything checks out, thus indeed \(\tan 15^\circ = 2 - \sqrt{3}\).

Key Concepts

Half-Angle Identity

The half-angle identity is a crucial tool in trigonometry, especially useful for breaking down angles into their components. When dealing with the identity for tangent, it helps us find the tangent of an angle that is half the size of a known angle. In our case, when calculating \( \tan 15^{\circ} \), we use the known \( \cos 30^{\circ} \), given as \( \frac{\sqrt{3}}{2} \), to find \( \tan 15^{\circ} = \tan \left( \frac{A}{2} \right) \). This identity is expressed as:

• \( \tan \left( \frac{A}{2} \right) = \frac{1 - \cos A}{\sin A} \)

For \( A = 30^{\circ} \), the identity simplifies our task and replaces complex geometric derivations, turning them into simple substitution and arithmetic operations.

Pythagorean Identity

Within trigonometry, the Pythagorean identity serves as a connection between sine and cosine functions of an angle \( A \). It is often employed to find a missing trigonometric function when one of them is already known. The identity is expressed as:

• \( \sin^2 A + \cos^2 A = 1 \)

By using this identity, when given \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \), we can find \( \sin 30^{\circ} \). Substituting the cosine value helps us determine:- \( \sin^2 30^{\circ} + \left( \frac{\sqrt{3}}{2} \right)^2 = 1 \)
- Simplifying this gives \( \sin^2 30^{\circ} = \frac{1}{4} \).
Thus, \( \sin 30^{\circ} = \frac{1}{2} \), completing our set of necessary values for further calculations.

Exact Values of Trigonometric Functions

Exact values in trigonometry refer to known specific values of trigonometric functions at common angles, often without decimal approximations. These are especially useful for eliminating the need for a calculator to find approximate trigonometric values. Some well-known values include:

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

Using these exact values enables precise computations and is essential in finding our initial condition for solving \( \tan 15^{\circ} \). Employing them allows exact results, increasing the reliability of further problem-solving steps.

Simplifying Trigonometric Expressions

Simplifying expressions is a vital skill in trigonometry, making complex expressions more manageable and often clearly revealing the solution. When deriving \( \tan 15^{\circ} \), simplification involved managing a complex expression into a straightforward difference. Steps included:

• Starting with \( \tan 15^{\circ} = \frac{1 - \cos 30^{\circ}}{\sin 30^{\circ}} \)
• Substituting known values to obtain \( \frac{1 - \frac{\sqrt{3}}{2}}{\frac{1}{2}} \)
• Rewriting and simplifying to \( 2 - \sqrt{3} \)

Through these simplifications, you not only execute operations efficiently but also check for accuracy, ensuring the solution is logically sound and mathematically correct. Such simplification enables even complex trigonometric problems to be handled with comfort and precision.