Problem 88 - Precalculus

Question

For the following exercises, find the exact value algebraically, and then confirm the answer with a calculator to the fourth decimal point. \(\tan \left(-15^{\circ}\right)\)

Step-by-Step Solution

Answer

\(\tan(-15^{\circ}) = -2 + \sqrt{3}\) which is approximately \(-0.2679\).

Step 1: Identify Trigonometric Identity

To find the exact value of \(\tan(-15^{\circ})\), let's use the tangent subtraction formula: \( \tan(a - b) = \frac{\tan(a) - \tan(b)}{1 + \tan(a)\tan(b)} \). Observe that \(-15^{\circ}\) can be expressed as \(45^{\circ} - 60^{\circ}\).

Step 2: Apply the Tangent Subtraction Formula

Using the tangent subtraction formula, substitute \(a = 45^{\circ}\) and \(b = 60^{\circ}\):\[ \tan(-15^{\circ}) = \frac{\tan(45^{\circ}) - \tan(60^{\circ})}{1 + \tan(45^{\circ})\tan(60^{\circ})} \]

Step 3: Substitute Known Values

Substitute \(\tan(45^{\circ}) = 1\) and \(\tan(60^{\circ}) = \sqrt{3}\) into the formula:\[ \tan(-15^{\circ}) = \frac{1 - \sqrt{3}}{1 + 1 \cdot \sqrt{3}} \]

Step 4: Simplify the Expression

Simplify the expression:\[ \tan(-15^{\circ}) = \frac{1 - \sqrt{3}}{1 + \sqrt{3}} \]

Step 5: Rationalize the Denominator

To simplify further, rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator, \(1 - \sqrt{3}\):\[ \tan(-15^{\circ}) = \frac{(1 - \sqrt{3})^2}{(1 + \sqrt{3})(1 - \sqrt{3})} \]

Step 6: Calculate the Exact Value

Solve the expression:\[ (1 - \sqrt{3})^2 = 1 - 2\sqrt{3} + 3 = 4 - 2\sqrt{3} \]\[ (1 + \sqrt{3})(1 - \sqrt{3}) = 1 - 3 = -2 \]So,\[ \tan(-15^{\circ}) = \frac{4 - 2\sqrt{3}}{-2} = -2 + \sqrt{3} \]

Step 7: Confirm with Calculator

Use a calculator to find \(\tan(-15^{\circ})\) and confirm it is approximately equal to \(-0.2679\).

Key Concepts

Trigonometric Identities

Trigonometric identities are essential tools in solving various mathematical problems, particularly those involving angles and trigonometric functions. They simplify complex trigonometric expressions and establish relationships between different trigonometric functions. For tangent, one of the core identities is the angle subtraction identity: \[ \tan(a - b) = \frac{\tan(a) - \tan(b)}{1 + \tan(a)\tan(b)} \] This identity allows us to find the tangent of an angle that can be expressed as a difference of two known angles.

• Use this identity when the given angle can be broken down into two known angles.
• It helps us in expressing complex angles in terms of familiar special angles like \(45^{\circ}, 30^{\circ}, 60^{\circ}\).

Understanding trigonometric identities helps build a strong foundation in trigonometry, which is crucial for success in higher mathematics.

Angle Subtraction

Angle subtraction is a key method used in trigonometry to solve for angles that may not correspond to standard or known angles on the unit circle. It involves breaking down a complex angle into the difference of two simpler angles. In our problem, \[ \tan(-15^{\circ}) = \tan(45^{\circ} - 60^{\circ}) \] By expressing \(-15^{\circ}\) as \(45^{\circ} - 60^{\circ}\), we make it easier to apply the tangent subtraction identity.

• This strategy utilizes known values of tan for \(45^{\circ}\) and \(60^{\circ}\).
• Helps compute exact values without the need for estimation or calculator use.

The angle subtraction technique not only simplifies calculations but also deepens your understanding of how different degrees relate on the trigonometric circle.

Rationalization

Rationalization is a process in mathematics used to eliminate radicals from the denominator of a fraction. This involves multiplying both the numerator and the denominator by the conjugate of the denominator. In the given exercise, we arrive at a fraction:\[ \frac{1 - \sqrt{3}}{1 + \sqrt{3}} \]To rationalize this, multiply by the conjugate \(1 - \sqrt{3}\):\[ \frac{(1 - \sqrt{3})(1 - \sqrt{3})}{(1 + \sqrt{3})(1 - \sqrt{3})} \]This simplifies the denominator using the difference of squares identity:\[ (1 + \sqrt{3})(1 - \sqrt{3}) = 1 - 3 = -2 \]The numerator becomes:\[ 1 - 2\sqrt{3} + 3 = 4 - 2\sqrt{3} \]Thus, \[ \tan(-15^{\circ}) = \frac{4 - 2\sqrt{3}}{-2} = -2 + \sqrt{3} \]

• Rationalization simplifies expressions to standard forms.
• Ensures calculations are clean and precise.
• This technique is frequently used in trigonometry to handle expressions with radicals.

Mastering rationalization allows you to manipulate and simplify mathematical expressions effectively.