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Chapter 6: Problem 33

Use the given information to determine the values of the remaining five trigonometric functions. (The angles are assumed to be acute angles. ) $$\tan A=\frac{\sqrt{2}-1}{\sqrt{2}+1}$$

Short Answer

Expert verified
Remaining trigonometric functions can be computed using identities and previously found \(\sin A\) and \(\cos A\).

Step by step solution

01

Simplify the Tangent Expression

We are given \(\tan A = \frac{\sqrt{2}-1}{\sqrt{2}+1}\). To simplify, multiply the numerator and the denominator by the conjugate of the denominator: \(\sqrt{2}-1\). \[\tan A = \frac{(\sqrt{2}-1)^2}{(\sqrt{2}+1)(\sqrt{2}-1)} = \frac{2 - 2\sqrt{2} + 1}{2 - 1} = 3 - 2\sqrt{2}. \]
02

Convert Tangent into Sine and Cosine

Since \(\tan A = \frac{\sin A}{\cos A}\), let's express \(\sin A\) and \(\cos A\) knowing that \(\tan A = 3 - 2\sqrt{2}\).Assume \(\sin A = 3 - 2\sqrt{2} \cdot \cos A \). We can now use trigonometric identities to solve for \(\sin A\) and \(\cos A\).
03

Use the Pythagorean Identity

Recall the Pythagorean identity: \(\sin^2 A + \cos^2 A = 1\). Substitute \(\sin A = (3 - 2\sqrt{2}) \cdot \cos A\) in the identity: \[((3 - 2\sqrt{2}) \cdot \cos A)^2 + \cos^2 A = 1.\]Simplify to find \(\cos A\).
04

Solve for \(\cos A\)

Expanding and solving the equation:\[(9 - 12\sqrt{2} + 8) \cdot \cos^2 A + \cos^2 A = 1, \]\[17\cos^2 A - 12\sqrt{2}\cos^2 A = 1,\] solve for \(\cos A\). Then, use \(\cos A\) to find \(\sin A\) using \(\sin A = (3 - 2\sqrt{2}) \cdot \cos A\).
05

Calculate Remaining Trigonometric Functions

Once \(\sin A\) and \(\cos A\) are known, calculate the remaining trigonometric functions:- \(\csc A = \frac{1}{\sin A}\)- \(\sec A = \frac{1}{\cos A}\)- \(\cot A = \frac{1}{\tan A}\)

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Key Concepts

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

Simplifying Expressions
Simplifying trigonometric expressions can help make calculations more manageable. In the original exercise, we simplified \(\tan A = \frac{\sqrt{2} - 1}{\sqrt{2} + 1}\) by multiplying both the numerator and the denominator by the conjugate of the denominator, \(\sqrt{2} - 1\). This technique is used to remove roots from the denominator and is useful in various math problems.
  • Multiply by conjugates to clear fractions in denominators.
  • Apply algebraic identities like \((a - b)^2 = a^2 - 2ab + b^2\) to expand expressions.
After applying these simplification techniques, we arrived at \(3 - 2\sqrt{2}\), which is a form that's easier to work with in further calculations.
Trigonometric Identities
Trigonometric identities are key tools in solving trigonometric equations. They provide relationships between the different trigonometric functions. The fundamental identities that are frequently used include:
  • Reciprocal identities: \(\csc A = \frac{1}{\sin A}\), \(\sec A = \frac{1}{\cos A}\), \(\cot A = \frac{1}{\tan A}\).
  • Tangent identity: \(\tan A = \frac{\sin A}{\cos A}\).
In this particular problem, we use trigonometric identities to express \(\tan A\) in terms of \(\sin A\) and \(\cos A\). These identities help us to convert one function into another, facilitating easier problem-solving.
Pythagorean Identity
The Pythagorean identity is a crucial identity in trigonometry, which states that \(\sin^2 A + \cos^2 A = 1\). This identity arises from the Pythagorean theorem and applies to right-angled triangles.
Using the equation \(\sin A = (3 - 2\sqrt{2}) \cdot \cos A\) from the original problem, we substitute it in the Pythagorean identity:\[((3 - 2\sqrt{2}) \cdot \cos A)^2 + \cos^2 A = 1.\]
Expanding this equation allows us to solve for \(\cos A\), which is pivotal in determining the other trigonometric functions.
Tangent
The tangent function, represented as \(\tan A\), is defined as the ratio of the opposite side to the adjacent side in a right-angled triangle. Another way to express it is as the ratio of \(\sin A\) to \(\cos A\):
\[\tan A = \frac{\sin A}{\cos A}.\]
In our problem, we simplified tangent to calculate the remaining trigonometric functions. Once simplified, \(\tan A\) is more manageable and links other functions through identities, highlighting its vital role in trigonometry.
Sine and Cosine Relationships
Sine and cosine are the foundational building blocks of trigonometric functions. These functions describe the relationship between the angles and lengths of the sides of a right-angled triangle.
In trigonometry, they relate through identities like:
  • \(\tan A = \frac{\sin A}{\cos A}\)
  • The Pythagorean identity: \(\sin^2 A + \cos^2 A = 1\)
In our exercise, expressing \(\sin A\) as \((3 - 2\sqrt{2}) \cdot \cos A\) and applying the Pythagorean identity enabled the determination of both \(\sin A\) and \(\cos A\). These relationships underscore the interconnected nature of trigonometric functions, allowing us to move from one function to another seamlessly.

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Most popular questions from this chapter

Let $$\begin{aligned}S(\theta) &=\sin \theta \\\C(\theta) &=\cos \theta \\\L(x) &=\ln x \end{aligned} \quad 0^{\circ} \leq \theta \leq 360^{\circ}$$ What is the domain of the function \((L \circ S)(\theta) ?\)

This exercise is adapted from a problem that appears in the classic text \(A\) Treatise on Plane and Advanced Trigonometry, 7th ed., by E. W. Hobson (New York: Dover Publications, 1928 ). (The first edition of the book was Oublished by Cambridge University Press in \(1891 .)\) Given: \(A, B,\) and \(C\) are acute angles such that \(\cos A=\tan B \quad \cos B=\tan C \quad \cos C=\tan A\) Prove: \(\sin A=\sin B=\sin C=2 \sin 18^{\circ}\) Follow steps (a) through (e) to obtain this result. (a) In each of the three given equations, use the identity \(\tan \theta=(\sin \theta) /(\cos \theta)\) so that the equations contain only sines and cosines. (b) In each of the three equations obtained in part (a), square both sides. Then use the identity \(\sin ^{2} \theta=1-\cos ^{2} \theta\) so that each equation contains only the cosine function. (c) For ease in writing, replace \(\cos ^{2} A, \cos ^{2} B,\) and \(\cos ^{2} C\) by \(a, b,\) and \(c,\) respectively. Now you have a system of three equations in the three unknowns \(a, b,\) and \(c .\) Solve for \(a, b,\) and \(c\) (d) Using the results in part (c), show that $$\sin A=\sin B=\sin C=\sqrt{\frac{3-\sqrt{5}}{2}}$$ (e) From Exercise \(54(\mathrm{f})\) we know that \(\sin 18^{\circ}=\) \((\sqrt{5}-1) / 4 .\) Show that the expression obtained in part (d) is equal to twice this expression for \(\sin 18^{\circ}\) This completes the proof. (Use the fact that two nonnegative quantities are equal if and only if their squares are equal.)

Use the given information to determine the remaining five trigonometric values. $$\cos \theta=-3 / 5, \quad 180^{\circ}<\theta<270^{\circ}$$

Given that \(\frac{\sin \alpha}{\sin \beta}=p\) and \(\frac{\cos \alpha}{\cos \beta}=q,\) express tan \(\alpha\) and \(\tan \beta\) in terms of \(p\) and \(q .\) (Assume that \(\alpha\) and \(\beta\) are acute angles.)

Write in terms of sine and cosine and simplify expression. $$\frac{\cot ^{2} \theta}{\csc ^{2} \theta}+\frac{\tan ^{2} \theta}{\sec ^{2} \theta}$$
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