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Chapter 7: Problem 3

Answer each question. Which one of the following equations has solution \(\frac{3 \pi}{4} ?\) A. arctan \(1=x\) B. \(\arcsin \frac{\sqrt{2}}{2}=x\) C. \(\arccos \left(-\frac{\sqrt{2}}{2}\right)=x\)

Short Answer

Expert verified
The equation (C) \(\arccos \left(-\frac{\sqrt{2}}{2}\right)=x\) has the solution \(\frac{3 \pi}{4}\).

Step by step solution

01

Identify the Given Value

The solution is given as \(\frac{3\pi}{4}\). This angle needs to be used to check each equation.
02

Evaluate arc functions

Evaluate each possible equation by finding the arc function value: 1. For A: \(\arctan 1\)2. For B: \(\arcsin \frac{\sqrt{2}}{2}\)3. For C: \(\arccos \left(-\frac{\sqrt{2}}{2}\) in radians.
03

Calculation of Each Arc Function

1. \(\arctan 1 = \frac{\pi}{4} \) since \(\tan \frac{\pi}{4} = 1\).2. \(\arcsin \frac{\sqrt{2}}{2} = \frac{\pi}{4}\) since \(\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}\).3. \(\arccos \left(-\frac{\sqrt{2}}{2}\right) = \frac{3\pi}{4}\) since \(\cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2}\).
04

Identify the Matching Equation

From the previous calculations, only \(\arccos \left(-\frac{\sqrt{2}}{2}\right)\) results in \(\frac{3\pi}{4}\). Thus, the correct equation is (C).

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Key Concepts

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

arctan
The **arctan** function, also known as the inverse tangent, is used to find the angle whose tangent value is a given number. For instance, \(\text{arctan}(y) = x\) means that \(\tan(x) = y\). This function outputs angles typically in the range from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).Here are some key points about arctan:
  • It is the inverse function of \(\tan\).
  • The output is an angle in radians or degrees.
  • Common values include \(\text{arctan}(1) = \frac{\pi}{4}\).
In the given exercise, when we evaluated \(\text{arctan}(1)\), the result was \(\frac{\pi}{4}\). Thus, option A was not the solution since it did not match the given value \(\frac{3\text{\pi}}{4}\).
arcsin
The **arcsin** function, also known as the inverse sine, helps find the angle whose sine value is a specified number. Mathematically, \(\text{arcsin}(y) = x\) indicates that \(\text{sin}(x) = y\). The outputs given by this function range typically from \(-\frac{\text{\pi}}{2}\) to \(\frac{\text{\pi}}{2}\).Key points to understand arcsin:
  • It is the inverse of the \(\text{sin}\) function.
  • Outputs are angles measured in radians or degrees.
  • A common value is \(\text{arcsin}(\frac{\text{\text{\text{sqrt}}{2}}}{2}) = \frac{\text{\pi}}{4}\).
In our example, when we calculated \(\text{arcsin}(\frac{\text{\text{\text{sqrt}}{2}}}{2})\), it equaled \(\frac{\text{\pi}}{4}\), which did not match the target value of \(\frac{3\text{\pi}}{4}\). So, option B was not the correct answer.
arccos
The **arccos** function, or inverse cosine, is used to determine the angle whose cosine value is provided. Written as \(\text{arccos}(y) = x\), it means \(\text{cos}(x) = y\). The usual range of values for arccos outputs is from 0 to \(\text{\text{\pi}}\).Essential information about arccos:
  • It is the inverse function of \(\text{cos}\).
  • The output is always an angle, either in radians or degrees.
  • An example is \(\text{arccos}(-\frac{\text{\text{\text{sqrt}}{2}}}{2}) = \frac{3\text{\pi}}{4}\).
In our problem, when we evaluated \(\text{arccos}(-\frac{\text{\text{\text{sqrt}}{2}}}{2})\), it resulted in \(\frac{3\text{\pi}}{4}\), which matched the specified value. Hence, option C was the correct solution.

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Most popular questions from this chapter

Verify that each equation is an identity. $$\frac{1+\cos 2 x}{\sin 2 x}=\cot x$$

Verify that each equation is an identity. $$\frac{\sin (x+y)}{\cos (x-y)}=\frac{\cot x+\cot y}{1+\cot x \cot y}$$

Solve each problem. When a musical instrument creates a tone of \(110 \mathrm{Hz}\). it also creates tones at \(220,330,440,550,660, \ldots\) Hz. A small speaker cannot reproduce the \(110-\mathrm{Hz}\) vibration but it can reproduce the higher frequencies, which are 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. (Source: Benade, A.. Fundamentals of Musical Acoustics, Dover Publications.) (a) We can model this phenomenon using a graphing calculator. In the window \([0,0.03]\) by \([-1,1],\) graph the upper harmonics represented by the pressure $$ P=\frac{1}{2} \sin [2 \pi(220) t]+\frac{1}{3} \sin [2 \pi(330) t]+\frac{1}{4} \sin [2 \pi(440) t] $$ (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.

(This discussion applies to functions of both angles and real numbers.) Consider the following. $$\begin{aligned}\cos (&\left.180^{\circ}-\theta\right) \\\&=\cos 180^{\circ} \cos \theta+\sin 180^{\circ} \sin \theta \\\&=(-1) \cos \theta+(0) \sin \theta \\\&=-\cos \theta\end{aligned}$$ \(\cos \left(180^{\circ}-\theta\right)=-\cos \theta\) is an example of a reduction formula, which is an identity that reduces a function of a quadrantal angle plus or minus \(\theta\) to a function of \(\theta\) alone. Another example of a reduction formula is \(\cos \left(270^{\circ}+\theta\right)=\sin \theta\) Here is an interesting method for quickly determining a reduction formula for a trigonometric function \(f\) of the form \(f(Q \pm \theta),\) where \(Q\) is a quadrantal angle. There are two cases to consider, and in each case, think of \(\boldsymbol{\theta}\) as a small positive angle in order to determine the quadrant in which \(Q \pm \theta\) will lie. Case 1 Suppose that \(Q\) is a quadrantal angle whose terminal side lies along the \(x\) -axis. Determine the quadrant in which \(Q \pm \theta\) will lie for a small positive angle \(\theta .\) If the given function \(f\) is positive in that quadrant, use a \(+\) sign on the reduced form. If \(f\) is negative in that quadrant, use a - sign. The reduced form will have that \(\operatorname{sign}, f\) as the function, and \(\theta\) as the argument. For example: CAN'T COPY THE GRAPH Case 2 Suppose that \(Q\) is a quadrantal angle whose terminal side lies along the \(y\) -axis. Determine the quadrant in which \(Q \pm \theta\) will lie for a small positive angle \(\theta .\) If the given function \(f\) is positive in that quadrant, use a \(+\) sign on the reduced form. If \(f\) is negative in that quadrant, use a - sign. The reduced form will have that sign, the cofunction of \(f\) as the function, and \(\theta\) as the argument. For example: CAN'T COPY THE GRAPH Use these ideas to write reduction formulas for each of the following. $$\tan \left(270^{\circ}-\theta\right)$$

Verify that each trigonometric equation is an identity. $$\sin ^{3} \theta+\cos ^{3} \theta=(\cos \theta+\sin \theta)(1-\cos \theta \sin \theta)$$
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