Problem 95 - A Graphical Approach to Precalculus with Limits

Question

Give the exact real number value of each expression. Do not use a calculator. $$\sin \left(\sin ^{-1} \frac{1}{2}+\tan ^{-1}(-3)\right)$$

Step-by-Step Solution

Answer

The exact value of the expression is \(\frac{1 - 3\sqrt{3}}{2\sqrt{10}}\).

Step 1: Understand Inverse Functions

The problem involves two inverse trigonometric functions: \(\sin^{-1}\) and \(\tan^{-1}\). These functions give angles as outputs. So \(\sin^{-1}\left(\frac{1}{2}\right)\) is the angle whose sine is \(\frac{1}{2}\), and \(\tan^{-1}(-3)\) is the angle whose tangent is \(-3\).

Step 2: Evaluate \(\sin^{-1}\left(\frac{1}{2}\right)\)

We know that \(\sin^{-1}\left(\frac{1}{2}\right)\) is an angle whose sine is \(\frac{1}{2}\). The common angle is \(\frac{\pi}{6}\) or \(30^\circ\) because \(\sin\frac{\pi}{6} = \frac{1}{2}\).

Step 3: Evaluate \(\tan^{-1}(-3)\)

For \(\tan^{-1}(-3)\), find an angle \(\theta\) such that \(\tan\theta = -3\). This is typically found by ideas of symmetry, knowing \(\tan(-\theta) = -\tan(\theta)\), which for \(\tan^{-1}(-3)\) is in the fourth quadrant. Denoting it as \(-\alpha\) where \(\tan(\alpha) = 3\). We denote it as \(\theta = \tan^{-1}(3)\), thus \(\tan^{-1}(-3) = -\theta\).

Step 4: Add the Angles

Add the angles from Step 2 and Step 3: \(\frac{\pi}{6} + \tan^{-1}(-3) = \frac{\pi}{6} - \theta\).

Step 5: Calculate the Final Expression

To solve \(\sin(\frac{\pi}{6} - \theta)\), use the sine subtraction formula: \(\sin(a - b) = \sin a \cos b - \cos a \sin b\). Substitute \(a = \frac{\pi}{6}\) and \(b = \theta\).

Step 6: Substitution and Simplification

Using the sine and cosine of \(\frac{\pi}{6}\): \(\sin\frac{\pi}{6} = \frac{1}{2}\) and \(\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}\). From \(\tan\theta = 3\), \(\sin\theta\) and \(\cos\theta\) can be derived from \(\frac{\sin\theta}{\cos\theta} = 3\), and \(\sin^2\theta + \cos^2\theta = 1\). This gives \(\cos^2\theta = \frac{1}{10}\) and \(\sin^2\theta = \frac{9}{10}\). Therefore, \(\cos\theta = \frac{1}{\sqrt{10}}\) and \(\sin\theta = \frac{3}{\sqrt{10}}\).

Step 7: Calculate \(\sin(\frac{\pi}{6} - \theta)\)

Substitute the known values for \(\sin\frac{\pi}{6}, \cos\frac{\pi}{6}, \sin\theta,\) and \(\cos\theta\) into the expression: \[\sin(\frac{\pi}{6} - \theta) = \left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{10}}\right) - \left(\frac{\sqrt{3}}{2}\right)\left(\frac{3}{\sqrt{10}}\right) = \frac{1}{2\sqrt{10}} - \frac{3\sqrt{3}}{2\sqrt{10}}\].

Step 8: Simplify the Expression

Combine the terms by expressing them under a common denominator: \(\frac{1 - 3\sqrt{3}}{2\sqrt{10}}\).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions, sometimes called arc functions, help us find angles from known trigonometric values. Commonly used inverse functions include \( \sin^{-1} \), \( \cos^{-1} \), and \( \tan^{-1} \). These functions essentially reverse trigonometric functions like sine, cosine, and tangent.

• For example, \( \sin^{-1}(x) \) returns the angle whose sine is \( x \).
• Similarly, \( \tan^{-1}(x) \) gives the angle whose tangent is \( x \).

These functions are useful because they allow us to calculate angles when given certain ratios. In our example, \( \sin^{-1}(\frac{1}{2}) \) returns \( \frac{\pi}{6} \) or \( 30^\circ \), as this is an angle whose sine is exactly \( \frac{1}{2} \). Similarly, \( \tan^{-1}(-3) \) instructs us to find an angle with a tangent of \(-3\).Further understanding involves identifying which quadrant the angle lies in, such as knowing \( \tan^{-1}(-3) \) is in the fourth quadrant.

Sine Subtraction Formula

The sine subtraction formula is an essential tool in trigonometry for dealing with expressions involving two angles. The formula is expressed as:
\[ \sin(a - b) = \sin a \cos b - \cos a \sin b \]
This formula is particularly helpful when evaluating expressions like \( \sin(\frac{\pi}{6} - \theta) \).

• We start by identifying each component: where \( a = \frac{\pi}{6} \) and \( b = \theta \).
• The task then is to find \( \sin a \), \( \cos b \) and then subtract \( \cos a \sin b \) accordingly.

Utilizing the formula helps us evaluate the specific trigonometric expression systematically. For example, knowing \( \sin\frac{\pi}{6} = \frac{1}{2} \) and \( \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2} \), we can substitute these values to simplify our expression further.

Angle Evaluation

Angle evaluation involves determining specific angle values using trigonometric ratios. When given a trigonometric function, evaluating the angle can help establish which values are being referenced.

• To solve \( \sin^{-1}(\frac{1}{2}) \), we identify that it refers to \( \frac{\pi}{6} \) or \( 30^\circ \).
• For \( \tan^{-1}(-3) \), angle symmetry helps understand it might be close to \(-\theta\) where \( \tan(\theta) = 3 \).

Angle evaluation is crucial because knowing exact angles allows us to use trigonometric identities accurately, ensuring all computations are correct. Practicing should help improve your comfort with approximating these angles.

Simplification of Expressions

Simplifying trigonometric expressions is often necessary to arrive at a final answer. It involves combining like terms and expressing terms under a common denominator.

• In the context of this exercise, the expression \( \sin(\frac{\pi}{6} - \theta) = \frac{1}{2\sqrt{10}} - \frac{3\sqrt{3}}{2\sqrt{10}} \) was simplified.
• This is done by combining terms to reach the simplest form: \( \frac{1 - 3\sqrt{3}}{2\sqrt{10}} \).

Simplification helps make calculations easier and answers neater. This step is crucial in achieving a precise and clear result. By practicing these operations, you'll gain a better understanding of manipulating trigonometric functions to solve for the unknown more efficiently.