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Chapter 7: Problem 45

Match each expression in Column I with its value in Column II. $$\mathbf{I}$$ $$\tan 15^{\circ}$$ $$\mathbf{II}$$ A. \(\frac{\sqrt{6}+\sqrt{2}}{4}\) B. \(\frac{-\sqrt{6}-\sqrt{2}}{4}\) C. \(\frac{\sqrt{6}-\sqrt{2}}{4}\) D. \(2+\sqrt{3}\) E. \(2-\sqrt{3}\) F. \(-2-\sqrt{3}\)

Short Answer

Expert verified
\( \tan 15^{\circ} = 2 - \sqrt{3} \) corresponding to Column II, option E.

Step by step solution

01

- Recall the identity for \tan(a - b)

Recall the tangent subtraction formula: \[\tan(a - b) = \frac{\tan(a) - \tan(b)}{1 + \tan(a) \tan(b)}\] We can use this formula to calculate \tan 15^{\backslashcirc} = \tan (45^{\backslashcirc} - 30^{\backslashcirc})
02

- Use known values of \tan 45^{\backslashcirc} and \tan 30^{\backslashcirc}

We know that \[ \tan 45^{\backslashcirc} = 1 \] and \[ \tan 30^{\backslashcirc} = \frac{1}{\backslashsqrt{3}}. \]
03

- Substitute values into the formula

Substitute these values into the tangent subtraction formula: \[ \tan 15^{\backslashcirc} = \frac{1 - \frac{1}{\backslashsqrt{3}}}{1 + 1 \backslashcdot \frac{1}{\backslashsqrt{3}}} \]
04

- Simplify the numerator and the denominator

Simplify the expression: \[ \frac{1 - \frac{1}{\backslashsqrt{3}}}{1 + \frac{1}{\backslashsqrt{3}}} = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}} {\frac{\sqrt{3} + 1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}. \]
05

- Simplify the resulting fraction

To further simplify, multiply the numerator and the denominator by \( \sqrt{3} - 1 \): \[ \frac{(\sqrt{3} - 1)^2}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{3 - 2\sqrt{3} +1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}. \]
06

- Match value with Column II

The simplified expression matches option E. Thus, \[ \tan 15^{\circ} = 2 - \sqrt{3}. \]

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Key Concepts

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

tangent subtraction formula
The tangent subtraction formula is a critical tool in trigonometry for calculating the tangent of the difference between two angles.
The formula is: \(\tan(a - b) = \frac{\tan(a) - \tan(b)}{1 + \tan(a) \tan(b)}\)
In our exercise, we need to find the value of \(\tan 15^{\circ}\). By recognizing that \(15^{\circ}\) can be expressed as \(45^{\circ} - 30^{\circ}\), we can apply the tangent subtraction formula.
This approach allows us to use known values of angles that are easier to handle.
simplification of expressions
Simplifying expressions is a foundational skill in algebra and trigonometry. In this exercise, after substituting the known tangent values into the tangent subtraction formula, our goal is to simplify the fraction:
\(\tan 15^{\circ} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}\)
Simplifying further involves rationalizing the numerator and the denominator:
\(\frac{(\sqrt{3} - 1)^2}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} \)
trigonometric values of special angles
Knowing the trigonometric values of special angles is essential for solving problems like these.
Special angles such as \(30^{\circ}\), \(45^{\circ}\), and \(60^{\circ}\) have well-known tangent values:
\(\tan 45^{\circ} = 1\) and \(\tan 30^{\circ} = \frac{1}{\sqrt{3}}\).
By using these values in the tangent subtraction formula, we can accurately determine the tangent of other angles, such as \(15^{\circ}\).
combining fractions
Combining fractions involves operations that make a complex expression easier to manage.
In the exercise, we started with:
\(\frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}\).
To simplify, we rationalized the expression by multiplying both the numerator and the denominator by the conjugate:
\(\sqrt{3} - 1\).
This yields a more straightforward form.
Multiplying out both parts and simplifying further achieved the final result of:
\(2 - \sqrt{3} \)

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Most popular questions from this chapter

Alternating Electric Current In the study of alternating electric current, instantaneous voltage is modeled by E=E_{\max } \sin 2 \pi f t where \(f\) is the number of cycles per second, \(E_{\max }\) is the maximum voltage, and \(t\) is time in seconds. (a) Solve the equation for \(t\) (b) Find the least positive value of \(t\) if \(E_{\max }=12, E=5,\) and \(f=100 .\) Use a calculator.

Solve each problem. Musicians sometimes tune instruments by playing the same tone on two different instruments and listening for a phenomenon known as beats. Beats occur when two tones vary in frequency by only a few hertz. When the two instruments are in tune, the beats disappear. The ear hears beats because the pressure slowly rises and falls as a result of this slight variation in the frequency. This phenomenon can be seen using a graphing calculator. (Source: Pierce, \(\mathrm{J}\)., The Science of Musical Sound, Scientific American Books.) (a) Consider the two tones with frequencies of \(220 \mathrm{Hz}\) and \(223 \mathrm{Hz}\) and pressures \(P_{1}=0.005 \sin 440 \pi t\) and \(P_{2}=0.005\) sin \(446 \pi t,\) respectively. Graph the pressure \(P=P_{1}+P_{2}\) felt by an eardrum over the 1 -sec interval \([0.15,1.15] .\) How many beats are there in 1 sec? (b) Repeat part (a) with frequencies of 220 and \(216 \mathrm{Hz}\) (c) Determine a simple way to find the number of beats per second if the frequency of each tone is given.

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