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If \(x=4 \tan \theta,\) express \(\sin \theta\) in terms of \(x\)

Short Answer

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Question: Express \(\sin \theta\) in terms of \(x\) given the relationship \(x = 4\tan \theta\). Answer: \(\sin\theta = \pm\frac{x}{\sqrt{16+x^2}}.\)

Step by step solution

01

Express \(\tan\theta\) in terms of \(x\)

We are given that \(x = 4\tan \theta\). First, we will express \(\tan\theta\) in terms of \(x\) by dividing both sides by 4: $$\tan \theta = \frac{x}{4}.$$
02

Express \(\sin\theta\) in terms of \(\tan\theta\)

Recall that \(\tan \theta = \frac{\sin\theta}{\cos\theta}\). We can solve for \(\sin\theta\) by multiplying both sides by \(\cos\theta\): $$\sin\theta = \tan\theta\cos\theta.$$
03

Find \(\cos\theta\) in terms of \(\tan\theta\)

Since \(\sin^2\theta + \cos^2\theta = 1\), we can express \(\cos^2\theta\) as \(1 - \sin^2\theta\). Now, recall that \(\tan^2\theta = \frac{\sin^2\theta}{\cos^2\theta}\). So, we can write $$\cos^2\theta = \frac{\sin^2\theta}{\tan^2\theta}.$$ Substitute \(1 - \sin^2\theta\) for \(\cos^2\theta\): $$1 - \sin^2\theta = \frac{\sin^2\theta}{\tan^2\theta}.$$ Multiply both sides by \(\tan^2\theta\) to get rid of the fraction: $$\tan^2\theta - \tan^2\theta\sin^2\theta = \sin^2\theta.$$
04

Express \(\sin\theta\) in terms of \(x\)

Now, we know $$\sin\theta = \tan\theta\cos\theta = \frac{x}{4}\cos\theta.$$ Replace \(\tan\theta\) with \(\frac{x}{4}\) in the expression we derived in step 3: $$\left(\frac{x}{4}\right)^2 - \left(\frac{x}{4}\right)^2\sin^2\theta = \sin^2\theta.$$ Simplify and solve for \(\sin^2\theta\): $$\frac{x^2}{16} - \frac{x^2}{16}\sin^2\theta = \sin^2\theta.$$ Factor out \(\sin^2\theta\) on the left side: $$\sin^2\theta\left(1+\frac{x^2}{16}\right) = \frac{x^2}{16}.$$ Now, divide both sides by \(\left(1+\frac{x^2}{16}\right)\): $$\sin^2\theta = \frac{\frac{x^2}{16}}{1+\frac{x^2}{16}}.$$ Finally, to find \(\sin\theta\), take the square root of both sides: $$\sin\theta = \pm\frac{x}{\sqrt{16+x^2}}.$$

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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, represented as \( \tan(\theta) \), is one of the primary trigonometric functions. It is defined in the context of a right triangle as the ratio of the length of the opposite side to the adjacent side. This is why we often see the formula \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \), derived from the sine function and cosine function.

In trigonometry, \( \theta \) usually represents an angle, and the values provided to \( \tan \) are typically angular measures in either degrees or radians.

The tangent function is unique because it doesn't restrict values between -1 and 1 like sine or cosine. Instead, its range is all real numbers since it can approach infinity as \( \theta \) approaches 90 degrees in a right triangle setup.

When solving problems like the given exercise where \( x = 4 \tan(\theta) \), it's common to first express \( \tan(\theta) \) in terms of \( x \). This involves simple algebraic manipulation where we express \( \tan(\theta) \) as \( \frac{x}{4} \).

Knowing the tangent function's relationship to the sine function allows the transformation between different trigonometric forms, creating formulas needed to solve complex trigonometric identities.
Sine Function
The sine function, denoted as \( \sin(\theta) \), is another fundamental function in trigonometry. It measures the ratio of the length of the side of a right triangle opposite the angle \( \theta \) to the length of the hypotenuse. As an angular function, sine helps us understand circular and periodic phenomena, particularly in physics and engineering.

The range of the sine function is between -1 and 1, and it only results in periodic outputs which makes it beneficial in analyzing cycles. Trigonometry uses this consistent functionality to solve equations that deal with periodicity and wave patterns.

In the exercise, the aim was to find \( \sin(\theta) \) in terms of \( x \). Through trigonometric identities, such as \( \sin^2(\theta) + \cos^2(\theta) = 1 \), and the relationship \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \), you can derive \( \sin(\theta) = \pm\frac{x}{\sqrt{16+x^2}} \). These steps are crucial, as they link \( \tan(\theta) \) and \( \sin(\theta) \) through algebra and trigonometric identities.

These identities reveal the importance of understanding the broader interactions across different trigonometric functions to solve problems effectively.
Trigonometric Ratios
Trigonometric ratios form the backbone of basic trigonometry, encapsulating relationships between the angles and sides of right-angled triangles. The primary trigonometric ratios are sine, cosine, and tangent, defined as follows:
  • Sine: \( \sin(\theta) = \frac{\text{Opposite Side}}{\text{Hypotenuse}} \)
  • Cosine: \( \cos(\theta) = \frac{\text{Adjacent Side}}{\text{Hypotenuse}} \)
  • Tangent: \( \tan(\theta) = \frac{\text{Opposite Side}}{\text{Adjacent Side}} \)

These functions allow the transformation of geometrical problems into algebraic ones, making calculations more manageable in many cases.

For example, using these ratios, you can solve problems where not all side lengths or angles are known by exploiting relationships like \( \sin^2(\theta) + \cos^2(\theta) = 1 \), a fundamental identity derived from the Pythagorean theorem.

Understanding these ratios is essential for more advanced studies and applications in calculus, physics, and engineering, where they help in modeling oscillatory behaviors, among other phenomena. Their applications in the problem shown involve expressing one trigonometric function in terms of another, simplifying calculations and enabling the solving of an otherwise more complex equation.

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Most popular questions from this chapter

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