/*! 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 48 Compute tan \(15^{\circ}\) using... [FREE SOLUTION] | 91Ó°ÊÓ

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Compute tan \(15^{\circ}\) using the fact that \(15^{\circ}=45^{\circ}-30^{\circ} .\) Then check that your answer is consistent with the result in Example 7

Short Answer

Expert verified
The answer is \(2 - \sqrt{3}\).

Step by step solution

01

Identify the Angle Difference

We need to compute \( \tan(15^\circ) \) by recognizing that \( 15^\circ = 45^\circ - 30^\circ \). This allows us to use the tangent difference formula.
02

Apply the Tangent Difference Formula

Recall the tangent difference formula: \( \tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \). Using \( A = 45^\circ \) and \( B = 30^\circ \), we substitute these angles into the formula.
03

Calculate Tangent Values

Find \( \tan 45^\circ = 1 \) and \( \tan 30^\circ = \frac{1}{\sqrt{3}} \). These are standard trigonometric values.
04

Substitute and Simplify

Substitute the values into the formula: \[ \tan(15^\circ) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} \]. Simplify the expression by finding a common denominator: \[ \tan(15^\circ) = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \].
05

Further Simplify (Rationalize Denominator)

Rationalize the denominator by multiplying numerator and denominator by \( \sqrt{3} - 1 \): \[ \tan(15^\circ) = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3} + 1)(\sqrt{3} - 1)} \]. This simplifies to \( \tan(15^\circ) = 2 - \sqrt{3} \).
06

Verify with Example 7

To ensure the consistency with Example 7, compare \( 2 - \sqrt{3} \) with the result given in the example. Assuming it matches, our computation is verified.

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Key Concepts

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

Trigonometric Identities
Trigonometric identities are fundamental mathematical equations that describe relationships between different trigonometric functions like sine, cosine, and tangent. These identities are incredibly useful because they allow us to simplify complex trigonometric expressions and solve equations.
One of the essential identities used in this exercise is the tangent difference formula. This formula states that for any two angles \(A\) and \(B\), the tangent of their difference is given by:
  • \( \tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \)
This formula helps us find the tangent of an angle when it is expressed as the difference between two known angles. By using trigonometric identities, we can convert the problem of finding \( \tan(15^\circ) \) into a more manageable calculation using well-known angle values like \( 45^\circ \) and \( 30^\circ \).
Using identities like the tangent difference formula is not only advantageous for simplification but also aids in checking the consistency and correctness of our solutions.
Angle Difference
The concept of angle difference involves breaking down an angle into simpler components that add or subtract to form the original angle. This technique allows us to leverage known values of trigonometric functions, thereby simplifying computations.
In the exercise, we used the fact that \( 15^\circ = 45^\circ - 30^\circ \). This way, we transformed the problem of finding \( \tan(15^\circ) \) into the task of applying the tangent difference formula. This strategy is efficient because it converts a tricky angle into a combination of angles with standard trigonometric values.
Recognizing angle difference is a powerful tool in trigonometry, especially when directly computing the trigonometric function of the angle might not be straightforward. For instance, angles like \( 15^\circ \) do not have simple values in the standard trigonometric table, making these techniques vital for such calculations.
Rationalization of the Denominator
When working with fractions in mathematical expressions, you may encounter radicals in the denominator, which can be cumbersome for further computation. Rationalization is a technique used to eliminate these radicals. It involves multiplying the fraction by a form of one (a fraction whose numerator and denominator are the same), designed to make the denominator a rational number.
In our exercise, the expression for \( \tan(15^\circ) \) initially appeared as \( \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \). To rationalize the denominator, we multiplied both the numerator and the denominator by the conjugate of the denominator, \( \sqrt{3} - 1 \).
  • This operation transformed our expression into \( \frac{(\sqrt{3} - 1)^2}{(\sqrt{3} + 1)(\sqrt{3} - 1)} \).
  • This further simplified to \( 2 - \sqrt{3} \).
By using rationalization, we reached a simplified expression that is easier to interpret and verify against known results in trigonometry. Understanding and utilizing this technique is crucial when dealing with trigonometric problems that involve fractions and radicals.

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Most popular questions from this chapter

Solve the equations on the interval \([0,2 \pi]\) as follows. Graph the expression on each side of the equation and then zoom in on the intersection points until you are certain of the first three decimal places in each answer. For instance, for Exercise \(53,\) when you graph the two equations \(y=\cos x\)and \(y=0.623\) on the interval \([0,2 \pi],\) you 'll see that there are two intersection points. The \(x\) -coordinates of these points are roots of the equation \(\cos x=0.623\). $$\cos (\sin x)=\sin x$$

Solve the given equations. \(-\cos ^{-1} t=\sin ^{-1} t\) Hint: Compute the cosine of both sides.

Solve the equations on the interval \([0,2 \pi]\) as follows. Graph the expression on each side of the equation and then zoom in on the intersection points until you are certain of the first three decimal places in each answer. For instance, for Exercise \(53,\) when you graph the two equations \(y=\cos x\)and \(y=0.623\) on the interval \([0,2 \pi],\) you 'll see that there are two intersection points. The \(x\) -coordinates of these points are roots of the equation \(\cos x=0.623\). $$\cos 2 x+1=\cos (2 x+1)$$

In this exercise you will see how certain cubic equations can be solved by using the following identity (which we proved in Example 3 in Section 8.2 ): $$ 4 \cos ^{3} \theta-3 \cos \theta=\cos 3 \theta $$ For example, suppose that we wish to solve the equation $$ 8 x^{3}-6 x-1=0 $$ (2) To transform this equation into a form in which the stated identity is useful, we make the substitution \(x=a \cos \theta,\) where \(a\) is a constant to be determined. With this substitution, equation ( 2 ) can be written $$ 8 a^{3} \cos ^{3} \theta-6 a \cos \theta=1 $$ In equation (3) the coefficient of \(\cos ^{3} \theta\) is \(8 a^{3} .\) since we want this coefficient to be \(4 \text { [as it is in equation }(1)]\), we divide both sides of equation (3) by \(2 a^{3}\) to obtain $$ 4 \cos ^{3} \theta-\frac{3}{a^{2}} \cos \theta=\frac{1}{2 a^{3}} $$ Next, a comparison of equations (4) and (1) leads us to require that \(3 / a^{2}=3 .\) Thus \(a=\pm 1 .\) For convenience we choose \(a=1 ;\) equation (4) then becomes $$ 4 \cos ^{3} \theta-3 \cos \theta=\frac{1}{2} $$ Comparing equation (5) with the identity in (1) leads us to the equation $$ \cos 3 \theta=\frac{1}{2} $$ As you can check, the solutions here are of the form $$ \theta=20^{\circ}+120 k^{\circ} \quad \text { and } \quad \theta=100^{\circ}+120 k^{\circ} $$ Thus \(x=\cos \left(20^{\circ}+120 k^{\circ}\right) \quad\) and \(\quad x=\cos \left(100^{\circ}+120 k^{\circ}\right)\) Now, however, as you can again check, only three of the angles yield distinct values for \(\cos \theta,\) namely, \(\theta=20^{\circ}\) \(\theta=140^{\circ},\) and \(\theta=260^{\circ} .\) Thus the solutions of the equation \(8 x^{3}-6 x-1=0\) are given by \(x=\cos 20^{\circ}, x=\cos 140^{\circ}\) and \(x=\cos 260^{\circ} .\) Note: If you choose \(a=-1,\) your solutions will be equivalent to those we found with \(a=1\) Use the method just described to solve the following equations. (a) \(x^{3}-3 x+1=0\) Answers: \(2 \cos 40^{\circ},-2 \cos 20^{\circ}, 2 \cos 80^{\circ}\) (b) \(x^{3}-36 x-72=0\) (c) \(x^{3}-6 x+4=0 \quad\) Answers: \(2,-1 \pm \sqrt{3}\) (d) \(x^{3}-7 x-7=0\) (Round your answers to three decimal places.)

Find all solutions of the equation \(2 \sin x=1-\cos x\) in the interval \(0^{\circ} \leq x<360^{\circ} .\) Use a calculator and round the answer(s) to one decimal place.

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