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Chapter 8: Problem 1

Use the given information to evaluate each expression. \(\cos \varphi=7 / 25 \quad\left(0_{i}<\varphi<90_{i}\right)\) (a) \(\sin 2 \varphi\) (b) \(\cos 2 \varphi\) (c) \(\tan 2 \varphi\)

Short Answer

Expert verified
(a) \( \frac{336}{625} \), (b) \( -\frac{527}{625} \), (c) \( -\frac{336}{527} \)

Step by step solution

01

Determine the sine of 蠁

Given that \( \cos \varphi = \frac{7}{25} \), use the Pythagorean identity \( \sin^2 \varphi + \cos^2 \varphi = 1 \) to find \( \sin \varphi \). Start by calculating \( \sin^2 \varphi = 1 - \cos^2 \varphi = 1 - \left( \frac{7}{25} \right)^2 = 1 - \frac{49}{625} = \frac{576}{625} \). Thus, \( \sin \varphi = \sqrt{\frac{576}{625}} = \frac{24}{25} \).
02

Calculate \\(ackslashackslash(sin 2 \\varphi\\)ackslash\\)

Use the double angle identity for sine: \( \sin 2 \varphi = 2 \sin \varphi \cos \varphi \). Substituting the values, we have \( \sin 2 \varphi = 2 \cdot \frac{24}{25} \cdot \frac{7}{25} = \frac{336}{625} \).
03

Calculate \\(ackslashackslash(cos 2 \\varphi\\)ackslash\\)

Use the double angle identity for cosine: \( \cos 2 \varphi = 2 \cos^2 \varphi - 1 = 2 \left( \frac{7}{25} \right)^2 - 1 = 2 \cdot \frac{49}{625} - 1 = \frac{98}{625} - 1 = \frac{98}{625} - \frac{625}{625} = -\frac{527}{625} \).
04

Calculate \\(ackslashackslash(tan 2 \\varphi\\)ackslash\\)

Use the relation \( \tan 2 \varphi = \frac{\sin 2 \varphi}{\cos 2 \varphi} \). Substitute the calculated values \( \sin 2 \varphi = \frac{336}{625} \) and \( \cos 2 \varphi = -\frac{527}{625} \), giving \( \tan 2 \varphi = \frac{\frac{336}{625}}{-\frac{527}{625}} = -\frac{336}{527} \).

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Key Concepts

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

Pythagorean identity
The Pythagorean identity is a fundamental trigonometric principle that relates the squares of the sine and cosine of an angle. It is written as: \[ \sin^2 \varphi + \cos^2 \varphi = 1 \]This identity helps us calculate one function given the other. In our exercise, we used this identity to find the sine of \( \varphi \) when given \( \cos \varphi = \frac{7}{25} \). By rearranging the Pythagorean identity, we calculated:
  • \( \sin^2 \varphi = 1 - \cos^2 \varphi \)
  • Substituting \( \cos \varphi = \frac{7}{25} \), we have \( \sin^2 \varphi = 1 - \left(\frac{7}{25}\right)^2 \)
  • This simplifies to \( \sin^2 \varphi = \frac{576}{625} \), leading to \( \sin \varphi = \frac{24}{25} \)
Understanding and applying the Pythagorean identity is essential for solving problems involving trigonometric functions. It provides a convenient way to confirm the consistency of trigonometric values as they always satisfy this identity.
double angle identities
Double angle identities are trigonometric formulas that express functions of double angles (\(2\varphi\)) in terms of the functions of the original angle (\(\varphi\)). These identities are incredibly useful for solving trigonometric equations, simplifying expressions, and calculating exact values. The most common double angle identities are:
  • \( \sin 2\varphi = 2 \sin \varphi \cos \varphi \)
  • \( \cos 2\varphi = 2 \cos^2 \varphi - 1 \) or \( \cos 2\varphi = 1 - 2 \sin^2 \varphi \)
  • \( \tan 2\varphi = \frac{2 \tan \varphi}{1 - \tan^2 \varphi} \)
In this exercise, we evaluated \( \sin 2\varphi \), \( \cos 2\varphi \), and \( \tan 2\varphi \) using these identities:- For \( \sin 2\varphi \), \( \sin 2\varphi = 2 \times \frac{24}{25} \times \frac{7}{25} = \frac{336}{625} \)- For \( \cos 2\varphi \), using \( 2 \cos^2 \varphi - 1 \), we found \( \cos 2\varphi = -\frac{527}{625} \)- For \( \tan 2\varphi \), we used the ratio \( \frac{\sin 2\varphi}{\cos 2\varphi} \), giving us \( \tan 2\varphi = -\frac{336}{527} \)These identities are powerful tools for dealing with trigonometric equations involving double angles.
trigonometric functions
Trigonometric functions are mathematical functions that relate the angles of a triangle to the lengths of its sides. They are fundamental to trigonometry, a branch of mathematics dealing with angles, triangles, and periodic phenomena. The three primary trigonometric functions are:
  • Sine (\( \sin \)): Ratio of the opposite side to the hypotenuse in a right triangle.
  • Cosine (\( \cos \)): Ratio of the adjacent side to the hypotenuse.
  • Tangent (\( \tan \)): Ratio of the opposite side to the adjacent side, or \( \tan \varphi = \frac{\sin \varphi}{\cos \varphi} \).
These functions are periodic, meaning they repeat their values in a regular pattern. Sine and cosine functions have a period of \( 2\pi \), while tangent has a period of \( \pi \). Because of their periodic nature, these functions are useful for modeling cyclical phenomena such as waves and oscillations. They are the building blocks for understanding more complex trigonometric identities and transformations, such as the double angle identities utilized in our exercise, making them an essential part of mathematical studies.

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Most popular questions from this chapter

Prove that the given equations are identities. $$\sin 2 \theta=2 \sin ^{3} \theta \cos \theta+2 \sin \theta \cos ^{3} \theta$$

If \(\cos \theta=x-1\) and \(0<\theta<\pi / 2,\) express \(2 \theta-\cos 2 \theta\) as a function of \(x.\)

Calculation of \(\sin 18^{\circ}, \cos 18^{\circ},\) and \(\sin 3^{\circ}\) (a) Prove that \(\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta\) (b) Supply a reason for each statement. (i) \(\sin 36^{\circ}=\cos 54^{\circ}\) (ii) \(2 \sin 18^{\circ} \cos 18^{\circ}=4 \cos ^{3} 18^{\circ}-3 \cos 18^{\circ}\) (iii) \(2 \sin 18^{\circ}=4 \cos ^{2} 18^{\circ}-3\) (c) In equation (iii), replace \(\cos ^{2} 18^{\circ}\) by \(1-\sin ^{2} 18^{\circ}\) and then solve the resulting equation for \(\sin 18^{\circ} .\) Thus show that \(\sin 18_{i}=\frac{1}{4}(\sqrt{5}-1)\) (d) Show that \(\cos 18_{i}=\frac{1}{4} \sqrt{10+2 \sqrt{5}}\) (e) Show that \(\sin 3^{\circ}\) is equal to \(\frac{1}{16}[(\sqrt{5}-1)(\sqrt{6}+\sqrt{2})-2(\sqrt{3}-1) \sqrt{5+\sqrt{5}}]\) (f) Use your calculator to check the results in parts (c), (d), and (e).

$$\begin{aligned} &\text { If } \tan B=\frac{n \sin A \cos A}{1-n \sin ^{2} A}, \text { show that }\\\ &\tan (A-B)=(1-n) \tan A \end{aligned}$$

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.)
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