/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 96 Give the exact real number value... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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

Short Answer

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

Step by step solution

01

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 \( \tan( \theta - \phi ) \).
02

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, \( \cos(\phi) = \sqrt{1 - \left(-\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \).
03

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 \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} \).\( \tan(\phi) = \frac{\sin(\phi)}{\cos(\phi)} = \frac{-\frac{3}{5}}{\frac{4}{5}} = -\frac{3}{4} \).
04

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,\[\tan(\theta - \phi) = \frac{\frac{4 + 3\sqrt{3}}{4\sqrt{3}}}{\frac{4\sqrt{3} - 3}{4\sqrt{3}}} = \frac{4 + 3\sqrt{3}}{4\sqrt{3} - 3}\]
05

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 \( 1 + \frac{25\sqrt{3}}{39} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use a graphical method to solve each equation over the interval \([0,2 \pi) .\) Round values to the nearest thousandth. $$\sin 4 x+\sin 2 x=2 \cos x$$

Solve each equation over the interval \([0,2 \pi)\) $$\sin 2 x=2 \cos ^{2} x$$

Hearing Different Tones When a musical instrument creates a tone of \(110 \mathrm{Hz}\), it also creates tones at \(220,330\) \(440,550,660, \dots\) Hz. A small speaker cannot reproduce the \(110-\mathrm{Hz}\) vibration, but it can reproduce the higher frequencies, called the upper harmonics. The low tones can still be heard, because the speaker produces difference tones of the upper harmonics. The difference between consecutive frequencies is \(110 \mathrm{Hz}\), and this difference tone will be heard by a listener. We can model this phenomenon with a graphing calculator. (a) In the window \([0,0.03]\) by \([-1,1]\), graph the upper harmonics represented by the pressure $$ \begin{aligned} P=& \frac{1}{2} \sin [2 \pi(220) t]+\frac{1}{3} \sin [2 \pi(330) t] \\ &+\frac{1}{4} \sin [2 \pi(440) t] \end{aligned} $$ (b) Estimate all \(t\) -coordinates where \(P\) is maximum. (c) What does a person hear in addition to the frequencies of \(220,330,\) and \(440 \mathrm{Hz} ?\) (d) Graph the pressure produced by a speaker that can vibrate at \(110 \mathrm{Hz}\) and above in the window \([0,0.03]\) by \([-2,2]\) (Image can't copy)

Suppose you are solving a trigonometric equation to find solutions in \(\left[0^{\circ}, 360^{\circ}\right)\) and your work leads to $$ \frac{1}{3} \theta=45^{\circ}, 60^{\circ}, 75^{\circ}, 90^{\circ} $$ What are the corresponding values of \(\theta ?\)

Solve each equation in part (a) analytically over the interval \([0,2 \pi) .\) Then use a graph to solve each inequality in part (b). (a) \(\cos 2 x=\frac{\sqrt{3}}{2}\) (b) \(\cos 2 x>\frac{\sqrt{3}}{2}\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.