If \(\tan \theta=\frac{12}{5}\), then \(\cos \theta=\)

Short Answer

Expert verified
\(\cos \theta = \pm \frac{5}{13}\)

Step by step solution

01

Write down the Pythagorean identity for tangent and cosine

Recall the Pythagorean identity for sine, cosine, and tangent functions: \[\sin^2 \theta + \cos^2 \theta = 1\] Since we have the value of \(\tan \theta\), we can rewrite the Pythagorean identity using the tangent function: \[\tan^2 \theta + 1 = \frac{1}{\cos^2 \theta}\]
02

Substitute the given value of tangent

Replace the value of \(\tan \theta\) in the equation: \[\left(\frac{12}{5}\right)^2 + 1 = \frac{1}{\cos^2 \theta}\]
03

Solve the equation for \(\cos^2 \theta\)

Simplify the equation and solve for \(\cos^2 \theta\): \[\frac{144}{25} + 1 = \frac{1}{\cos^2 \theta}\] Convert 1 to a fraction with a denominator of 25: \[\frac{144}{25} + \frac{25}{25} = \frac{1}{\cos^2 \theta}\] Add the fractions on the left side: \[\frac{169}{25} = \frac{1}{\cos^2 \theta}\]
04

Take the reciprocal of both sides of the equation

To solve for \(\cos^2 \theta\), take the reciprocal of both sides of the equation: \[\frac{25}{169} = \cos^2 \theta\]
05

Find the value of \(\cos \theta\)

To find the value of \(\cos \theta\), take the square root of both sides of the equation, keeping in mind that the cosine function can have positive or negative values depending on the angle's quadrant: \[\pm \sqrt{\frac{25}{169}} = \cos \theta\] Simplify the square root: \[\pm \frac{5}{13} = \cos \theta\] To give a unique solution, state the quadrant in which the angle \(\theta\) lies, or mention what the range of the angle is. However, without that information, we find that the value of \(\cos \theta\) can be \(\frac{5}{13}\) or \(-\frac{5}{13}\).

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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 primary trigonometric functions and is abbreviated as \(\tan\). It relates the angles of a right-angled triangle to the ratio of the triangle's opposite side over its adjacent side. More formally, \(\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\).

In the unit circle, where all triangles have a hypotenuse of length 1, the tangent function represents the length of the segment tangent to the circle from the point where the line through the origin makes an angle \(\theta\) with the positive x-axis.

Special Values and Properties

Tangent has some unique properties. For example, it is periodic with a period of \(\pi\) radians, meaning \(\tan(\theta) = \tan(\theta + n\pi)\) for all integers \(n\). Additionally, it has asymptotes, points where the function heads towards infinity, at \(\theta = \frac{\pi}{2} + n\pi\). This is due to the cosine function, which is in the denominator of the tangent ratio, approaching zero at these points.
Cosine Function
The cosine function, denoted as \(\cos\), is another fundamental trigonometric function that represents the ratio of the adjacent side to the hypotenuse of a right-angled triangle. Simply put, \(\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}\).

Unit Circle Interpretation

In trigonometry, using the unit circle is a powerful way to understand the function. On this circle, the cosine function gives the x-coordinate of the point formed by the intersection of the terminal side of the angle \(\theta\) and the circle.

Regarding the shape of the cosine graph, it is a wave which starts at 1 (when \(\theta = 0\)) and oscillates between the values 1 and -1 for every period of \(2\pi\) radians. The cosine function is an even function, meaning it is symmetric with respect to the y-axis, which tells us \(\cos(\theta) = \cos(-\theta)\).
Trigonometric Identities
Trigonometric identities are equations that relate the trigonometric functions to one another. They are true for all values of the involved angles and are incredibly useful for simplifying expressions and solving equations that involve trigonometric functions.

Pythagorean Identity

The Pythagorean identity, as shown in the original exercise, states that \(\sin^2(\theta) + \cos^2(\theta) = 1\). Since the tangent function is \(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\), we can manipulate the Pythagorean identity to involve tangent as well: \(\tan^2(\theta) + 1 = \frac{1}{\cos^2(\theta)}\), a crucial step in solving the provided exercise.

Understanding and being able to derive these relationships between sinusoidal functions can considerably streamline the process of solving trigonometric problems.
Solving Trigonometric Equations
Solving trigonometric equations involves finding the values of the angles that satisfy the equation. The steps usually involve using trigonometric identities, algebraic manipulation, and sometimes the inverse trigonometric functions.

Methodology

Generally, one starts by isolating the trigonometric function, as was done in the steps of the provided solution. It's important to consider the function’s domain and range since functions like cosine and sine can have multiple solutions within a single period. Furthermore, the quadrant of the angle can affect the sign of the solution, as with the cosine function, which can take on both positive and negative values.

Always verify whether additional solutions exist by exploring the fundamental period of the trigonometric function involved. Lastly, it’s crucial to simplify results where possible and state them clearly, as ambiguous solutions can lead to confusion.

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Most popular questions from this chapter

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