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Find the exact value of each expression, if possible, without using a calculator. (a) \(\arccos (-\sqrt{3})\) (b) \(\arcsin \frac{\sqrt{2}}{2}\)

Short Answer

Expert verified
\(\arccos (-\sqrt{3}/2) = 5π/6\) and \(\arcsin (\sqrt{2}/2) = π/4\)

Step by step solution

01

Analyze the arccosine function

Arccos, or inverse cosine, gives us an angle whose cosine is the input value. The range for arccos is \(0 ≤ θ ≤ π\). Here, the input value is \(-\sqrt{3}/2\). The cosine of \(π/6\) is \(\sqrt{3}/2\) and of \(5π/6\) is \(-\sqrt{3}/2\). So, \(\arccos (-\sqrt{3}/2)\) will take the value of \(5π/6\).
02

Analyze the arcsine function

Arcsin, or inverse sine, gives us an angle whose sine is the input value. The range for arcsin is \(-π/2 ≤ θ ≤ π/2\). Here, the input value is \(\sqrt{2}/2\). The sine of both \(π/4\) and \(-π/4\) are \(\sqrt{2}/2\). However, since \(-π/4\) is not in the domain of arcsin, the only possibility left is \(π/4\). So, \(\arcsin (\sqrt{2}/2)\) will take the value of \(π/4\).

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Key Concepts

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

Arccosine Function
The arccosine function, often represented as \( \arccos \), is an inverse trigonometric function. It helps you find an angle from a given cosine value. If you have a cosine value, you can use \( \arccos \) to figure out which angle produced that cosine.

One thing to remember about \( \arccos \) is its range. It always results in an angle between 0 and \( \pi \) radians, which is between 0 and 180 degrees. This means, no matter the input, the output (the angle) will fall within this range. It helps provide a unique angle for any given cosine value within this range.
  • If the input is positive or zero, \( \arccos \) gives an angle on the upper half of the unit circle, meaning between 0 and 90 degrees.
  • If the input is negative, the angle is on the lower half, between 90 and 180 degrees.
Using this information, you can determine that for \( \arccos (-\sqrt{3}/2) \), the angle is \( 5\pi/6 \) because the cosine of this angle within \( 0 \) to \( \pi \) is \( -\sqrt{3}/2 \).
Arcsine Function
The arcsine function, written as \( \arcsin \), is another key inverse trigonometric function. It allows you to determine an angle when you already know the sine of that angle. Understanding \( \arcsin \) is crucial for working with trigonometric values in reverse.

The range of the \( \arcsin \) function is between \(-\pi/2\) and \(\pi/2\), which translates to angles from -90 to 90 degrees. This characteristic ensures that there is a one-to-one correspondence between sine values and angles.
  • For positive sine values, \( \arcsin \) gives an angle in the first quadrant (0 to 90 degrees).
  • For negative sine values, it produces angles in the fourth quadrant (-90 to 0 degrees).
In our case, \( \arcsin(\sqrt{2}/2) \) results in \( \pi/4 \), as the sine of \( \pi/4 \) is \( \sqrt{2}/2 \) and \( \pi/4 \) is within the \( \arcsin \) range.
Trigonometric Values
Trigonometric values relate to the sine and cosine of angles in a right-angled triangle or the unit circle. Essential understanding of these values can help you solve equations like those with the \( \arccos \) and \( \arcsin \) functions.

Using the unit circle, the most commonly referenced angles are \( 0, \pi/6, \pi/4, \pi/3, \pi/2, \) and their respective angles in radians such as \( \pi, \) and \( 2\pi \). Each of these angles has known sine and cosine values that are crucial when working with inverse trigonometric functions.
  • For \( \pi/6,\) the cosine is \( \sqrt{3}/2 \) and sine is \( 1/2\).
  • For \( \pi/4, \) both sine and cosine are \( \sqrt{2}/2 \).
  • For \( \pi/3, \) the sine is \( \sqrt{3}/2 \) and cosine is \( 1/2 \).
When solving for inverse functions like \( \arccos \) and \( \arcsin \), knowing these trigonometric values helps directly match angles to given trigonometric results. This knowledge allows you to work efficiently without always needing a calculator, especially within concepts revolving around inverse trigonometric functions.

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Most popular questions from this chapter

True or False Determine whether the statement is true or false. Justify your answer. An example of a bearing used in aviation is \(\mathrm{S} 25^{\circ} \mathrm{W}.\)

A 20 -meter line is used to tether a helium-filled balloon. Because of a breeze, the line makes an angle of approximately \(85^{\circ}\) with the ground. (a) Draw a right triangle that gives a visual representation of the problem. Show the known quantities of the triangle and use a variable to indicate the height of the balloon. (b) Use a trigonometric function to write an equation involving the unknown quantity. (c) What is the height of the balloon? (d) The breeze becomes stronger, and the angle the balloon makes with the ground decreases. How does this affect your triangle from part (a)? (e) Complete the table, which shows the heights (in meters) of the balloon for decreasing angle measures \(\theta\). $$\begin{array}{|l|l|l|l|l|}\hline \text { Angle, } \theta & 80^{\circ} & 70^{\circ} & 60^{\circ} & 50^{\circ} \\\\\hline \text { Height } & & & & \\\\\hline\end{array}$$ $$\begin{array}{|l|l|l|l|l|}\hline \text { Angle, } \theta & 40^{\circ} & 30^{\circ} & 20^{\circ} & 10^{\circ} \\\\\hline \text { Height } & & & & \\\\\hline\end{array}$$ (f) As the angle the balloon makes with the ground approaches \(0^{\circ},\) how does this affect the height of the balloon? Draw a right triangle to explain your reasoning.

Simplify the radical expression. \(\frac{5 \sqrt{5}}{2 \sqrt{10}}\)

Find the exact value of each function for the given angle for \(f(\theta)=\sin \theta\) and \(g(\theta)=\cos \theta .\) Do not use a calculator. (a) \((f+g)(\theta)\) (b) \((g-f)(\theta)\) (c) \([g(\theta)]^{2}\) (d) \((f g)(\theta)\) (e) \(f(2 \theta)\) (f) \(g(-\boldsymbol{\theta})\) $$\theta=5 \pi / 6$$

Determine whether the statement is true or false. Justify your answer. The graph of \(y=6-\frac{3}{4} \sin \frac{3 x}{10}\) has a period of \(\frac{20 \pi}{3}.\)

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