Problem 96 - A Graphical Approach to Precalculus with Limits

Question

Give the exact real number value of each expression. Do not use a calculator. $$\tan \left(\cos ^{-1} \frac{\sqrt{3}}{2}-\sin ^{-1}\left(-\frac{3}{5}\right)\right)$$

Step-by-Step Solution

Answer

The exact value is \(1 + \frac{25\sqrt{3}}{39}\).

Step 1: Understanding the Inverse Trigonometric Functions

Let's define the angles based on the given inverse trigonometric functions: - Let \( \theta = \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) \). This implies that \( \cos(\theta) = \frac{\sqrt{3}}{2} \).- Let \( \phi = \sin^{-1} \left( -\frac{3}{5} \right) \). This implies that \( \sin(\phi) = -\frac{3}{5} \). We need to find \( \tan( \theta - \phi ) \).

Step 2: Determine \( \sin(\theta) \) and \( \cos(\phi) \)

Since \( \theta = \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) \), and knowing \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \), we have \( \theta = 30^\circ \) or \( \theta = \frac{\pi}{6} \) radians. Therefore, \( \sin(\theta) = \sin(30^\circ) = \frac{1}{2} \).For \( \phi = \sin^{-1} \left( -\frac{3}{5} \right) \), we use the Pythagorean identity: \( \cos^2(\phi) = 1 - \sin^2(\phi) \). So, \( \cos(\phi) = \sqrt{1 - \left(-\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \).

Step 3: Use the Tangent Difference Identity

The tangent difference identity is \( \tan(\theta - \phi) = \frac{\tan(\theta) - \tan(\phi)}{1 + \tan(\theta) \tan(\phi)} \).We know \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} \).\( \tan(\phi) = \frac{\sin(\phi)}{\cos(\phi)} = \frac{-\frac{3}{5}}{\frac{4}{5}} = -\frac{3}{4} \).

Step 4: Calculate \( \tan(\theta - \phi) \)

Plugging the values into the tangent difference identity:\[\tan(\theta - \phi) = \frac{\frac{1}{\sqrt{3}} - \left(-\frac{3}{4}\right)}{1 + \left(\frac{1}{\sqrt{3}}\right) \left(-\frac{3}{4}\right)}\]\[= \frac{\frac{1}{\sqrt{3}} + \frac{3}{4}}{1 - \frac{3}{4\sqrt{3}}}\]To simplify, let's combine the fractions separately:\[\frac{1}{\sqrt{3}} + \frac{3}{4} = \frac{4 + 3\sqrt{3}}{4\sqrt{3}}\]\[1 - \frac{3}{4\sqrt{3}} = \frac{4\sqrt{3} - 3}{4\sqrt{3}}\]Thus,\[\tan(\theta - \phi) = \frac{\frac{4 + 3\sqrt{3}}{4\sqrt{3}}}{\frac{4\sqrt{3} - 3}{4\sqrt{3}}} = \frac{4 + 3\sqrt{3}}{4\sqrt{3} - 3}\]

Step 5: Simplify the Expression

Now simplify the fraction to obtain the exact value:\[\tan(\theta - \phi) = \frac{(4 + 3\sqrt{3})(4\sqrt{3} + 3)}{(4\sqrt{3} - 3)(4\sqrt{3} + 3)}\]Denote \( a = 4 + 3\sqrt{3} \) and expand:\[(a)(4\sqrt{3} + 3) = 4\sqrt{3}\cdot 4 + 4\cdot 3 + 3\sqrt{3}\cdot 4\sqrt{3} + 9 = 16\sqrt{3} + 12 + 27 + 9\sqrt{3} = 25\sqrt{3} + 39\]For the denominator using difference of squares, \((4\sqrt{3})^2 - 3^2 = 48 - 9 = 39\).Thus, \( \tan(\theta - \phi) = \frac{25\sqrt{3} + 39}{39} \). Reducing gives the final result of \( 1 + \frac{25\sqrt{3}}{39} \).

Key Concepts

Tangent Difference Identity

The tangent difference identity is a wonderful trigonometric identity that can simplify complex expressions involving differences of angles. It is expressed as:

• \( \tan(\theta - \phi) = \frac{\tan(\theta) - \tan(\phi)}{1 + \tan(\theta) \tan(\phi)} \)

When dealing with tangent expressions of two angles, the difference identity allows us to find the tangent of the difference by using the tangents of the individual angles. Remember, this identity depends on knowing the tangents of each angle beforehand.
In our problem, we have two angles derived from inverse trigonometric functions. We use the tangent difference identity to express \( \tan(\theta - \phi) \) as a fraction involving the tangent of each angle. This is crucial in avoiding calculator usage and sticking to exact values rather than approximations. Keep in mind, knowing this identity means you're one step closer to solving trigonometric problems with ease.

Exact Trigonometric Values

Exact trigonometric values are specific values typically found on the unit circle for common angles such as 30°, 45°, 60°, etc. Using these values instead of decimal approximations keeps our work precise. For instance:

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

In our problem, the angle \( \theta \) is derived from using the inverse cosine function, \( \cos^{-1}(\frac{\sqrt{3}}{2}) \), and can be linked directly to the angle 30°. This angle is crucial because it provides us the exact trigonometric values we need to solve the expression without a calculator.
Meanwhile, angle \( \phi \), derived from \( \sin^{-1}(-\frac{3}{5}) \), isn't a standard angle, but using the Pythagorean identity allows the determination of the related \( \cos(\phi) \). With both angles set in exact forms, the calculations remain pure, presenting exact results as opposed to rounded figures.

Pythagorean Identity

The Pythagorean identity is one of the fundamental trigonometric identities, crucial for expressing trigonometric functions in terms of each other. It is given as:

• \( \sin^2(\theta) + \cos^2(\theta) = 1 \)

In the problem, this identity becomes useful when determining \( \cos(\phi) \) from \( \sin(\phi) \). With \( \sin(\phi) = -\frac{3}{5} \), the Pythagorean identity helps us deduce that:

• \( \cos^2(\phi) = 1 - \sin^2(\phi) = 1 - \left(-\frac{3}{5}\right)^2 = \frac{16}{25} \)

This gives \( \cos(\phi) = \frac{4}{5} \) as \( \phi \) is in a principal domain where cosine remains positive.
Using these relationships, the complexities of inverse trigonometry can become manageable. Also, considering the identity ensures our work remains aligned with exact forms of trigonometric functions.