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Find all solutions in the interval \(0^{\circ} \leq \theta \leq 360^{\circ} .\) Where necessary, use a calculator and round to one decimal place. $$\tan ^{2} \theta-\tan \theta-1=0$$

Short Answer

Expert verified
The solutions are \( \theta = 58.0^\circ, 238.0^\circ, 148.0^\circ,\) and \(328.0^\circ \).

Step by step solution

01

Recognize the Equation Format

The equation given is a quadratic in terms of \(\tan \theta\). It is of the form \( ax^2 + bx + c = 0 \), where \(x = \tan \theta\), \(a = 1\), \(b = -1\), and \(c = -1\). This means we can directly apply the quadratic formula to solve it.
02

Apply the Quadratic Formula

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve for \(x = \tan \theta\). Here, \(a = 1\), \(b = -1\), and \(c = -1\). Substitute these values into the formula to get:\[x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}\]
03

Calculate the Roots

Calculate the exact roots: 1. \( \tan \theta = \frac{1 + \sqrt{5}}{2} \) 2. \( \tan \theta = \frac{1 - \sqrt{5}}{2} \)Using a calculator, \(\frac{1 + \sqrt{5}}{2} \approx 1.618\) and \(\frac{1 - \sqrt{5}}{2} \approx -0.618\).
04

Determine the Angles for Each Root

Now compute \( \theta \) for each value using \( \tan \theta \):1. \( \tan \theta = 1.618 \): Find \( \theta \) where \( \tan \theta = 1.618 \). Using a calculator, \( \theta \approx 58.0^\circ \). Since tangent has a period of \(180^\circ\), the next angle with the same tangent value is \(58.0^\circ + 180^\circ = 238.0^\circ\).2. \( \tan \theta = -0.618 \): Find \( \theta \) where \( \tan \theta = -0.618 \). Using a calculator, \( \theta \approx 328.0^\circ \). Again due to the periodicity of tangent, another solution is \(328.0^\circ - 180^\circ = 148.0^\circ\).
05

Verify Angle Solutions in Range

Ensure that all calculated angles \(58.0^\circ, 238.0^\circ, 328.0^\circ,\) and \(148.0^\circ\) are within the given interval \(0^\circ \leq \theta \leq 360^\circ\). Since all angles are in this interval, they are valid solutions.
06

Confirm and Summarize the Solutions

After verifying, the solutions for \(\theta\) that satisfy \(\tan^2 \theta - \tan \theta - 1 = 0\) within the given range are \( \theta = 58.0^\circ, 238.0^\circ, 148.0^\circ,\) and \(328.0^\circ \). These are the angles at which the tangent values solve the original equation.

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Key Concepts

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

Tangent Function
The tangent function is one of the basic trigonometric functions, fundamental to understanding trigonometric equations. In a right triangle, the tangent of an angle \( \theta \) represents the ratio of the opposite side to the adjacent side. This can be written as:
  • \( \tan \theta = \frac{\text{opposite}}{\text{adjacent}} \)
However, tangent can also be defined using the sine and cosine functions as:
  • \( \tan \theta = \frac{\sin \theta}{\cos \theta} \)
Tangent is unique because unlike sine and cosine, its value can potentially range from negative infinity to positive infinity. This is critical in solving trigonometric equations.
In this problem, the solution revolves around expressing \( \tan^2 \theta - \tan \theta - 1 = 0 \) as a quadratic equation. Here, \( \tan \theta \) acts as the variable, allowing us to use familiar algebraic techniques to solve it.
Quadratic Formula
The quadratic formula is a powerful tool used in algebra to find the roots of a quadratic equation. A quadratic equation typically takes the form \( ax^2 + bx + c = 0 \). In our trigonometric problem, the equation \( \tan^2 \theta - \tan \theta - 1 = 0 \) is treated as a quadratic with \( x = \tan \theta \). The coefficients are:
  • \( a = 1 \)
  • \( b = -1 \)
  • \( c = -1 \)
We substitute these into the quadratic formula:
  • \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
This gives:
  • \( x = \frac{1 \pm \sqrt{5}}{2} \)
After calculating, we obtain two possible values for \( \tan \theta \): approximately 1.618 and -0.618. These are the solutions that arise from the quadratic formula application, essential in identifying potential angles for \( \theta \).
Angle Measurement
Finding angles that match specific tangent values involves understanding angle measures. Here, we seek angles within the interval \( 0^\circ \leq \theta \leq 360^\circ \) where \( \tan \theta \) is approximately 1.618 and -0.618.
Using an inverse tangent function on calculators reveals:
  • For \( \tan \theta = 1.618 \), \( \theta \approx 58.0^\circ \).
  • For \( \tan \theta = -0.618 \), \( \theta \approx 328.0^\circ \).
These angle measurements guide us to solve for \( \theta \). It's crucial to remember that the tangent function can be positive or negative in different quadrants, affecting our calculated angles and necessitating the inclusion of complementary angles for a complete solution.
Periodicity of Functions
The periodicity of the tangent function means it repeats its values over regular intervals. Specifically, the tangent function has a period of \( 180^\circ \). This property is crucial when finding all solutions for our trigonometric equation within the interval \( 0^\circ \leq \theta \leq 360^\circ \).
Starting with \( \theta = 58.0^\circ \) for \( \tan \theta = 1.618 \), adding the period gives:
  • Another angle: \( 238.0^\circ \).
Similarly, for \( \tan \theta = -0.618 \) starting with \( \theta = 328.0^\circ \), subtracting the period results in:
  • An additional angle: \( 148.0^\circ \).
These angles reflect the multiple intersections of the tangent graph with specific values throughout its periods and are essential for capturing all possible solutions in the given range.

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Most popular questions from this chapter

Evaluate \(\sin \left(2 \tan ^{-1} 4\right) .\) Hint: \(\sin 2 \theta=2 \sin \theta \cos \theta.\)

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 \(\cos x=0.623\). $$\sin ^{3} x+\cos ^{3} x=0.5$$

In Example 10 we found that \(\sec \left(\tan ^{-1} x\right)=\sqrt{1+x^{2}}\) for \(x>0 .\) Actually, this identity is valid for all real numbers. Demonstrate this visually by graphing the two functions \(y=\sec \left(\tan ^{-1} x\right)\) and \(y=\sqrt{1+x^{2}}.\)

The results and the techniques are taken from the article by Zalman Usiskin, "Products of Sines," which appeared in The Two-Year College Mathematics Journal, vol. \(10(1979),\) pp. \(334-340\) (a) Use your calculator to check that \(\sin 18^{\circ} \sin 54^{\circ}=1 / 4\) (b) Supply reasons for each of the following steps to prove that the equation in part (a) is indeed correct. (i) \(\sin 72^{\circ}=2 \sin 36^{\circ} \cos 36^{\circ}=2 \sin 36^{\circ} \sin 54^{\circ}\) (ii) \(\sin 72_{i}=4 \sin 18_{i} \cos 18_{i} \sin 54_{i}\) \(=4 \sin 18 ; \sin 72 ; \sin 54_{i}\) (iii) \(1 / 4=\sin 18^{\circ} \sin 54^{\circ}\)

Use graphs to determine whether there are solutions for each equation in the interval \([0,1] .\) If there are solutions, use the graphing utility to find them accurately to two decimal places. (a) \(\arccos x=2 \sin 3 x\) (b) \(\arccos x=2 \sin 4 x\)

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