Problem 69 - Algebra and Trigonometry

Question

Find the values of the trigonometric functions of \(t\) from the given information. \(\tan t=-\frac{3}{4}, \quad \cos t>0\)

Step-by-Step Solution

Answer

\(\sin t = -\frac{3}{5}, \cos t = \frac{4}{5}\) and \(\tan t = -\frac{3}{4}\).

Step 1: Identify the Quadrant

Since \(\tan t = -\frac{3}{4}\) and \(\tan t\) is negative, we know that \(t\) is in either the second or fourth quadrant. Given \(\cos t > 0\), \(t\) must be in the fourth quadrant because cosine is positive there.

Step 2: Determine Sides of the Right Triangle

To determine \(\sin t\) using \(\tan t = -\frac{3}{4}\). The opposite side is 3, the adjacent side is 4, making the hypotenuse \(h\) determined by the Pythagorean theorem: \(h = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\).

Step 3: Calculate Trigonometric Functions

In the fourth quadrant, sine is negative. Thus, \(\sin t = -\frac{3}{5}\) and \(\cos t = \frac{4}{5}\) because cosine is positive in the fourth quadrant by definition. The original information, \(\tan t\), is verified as \(\tan t = \frac{\sin t}{\cos t} = \frac{-\frac{3}{5}}{\frac{4}{5}} = -\frac{3}{4}\).

Step 4: Check the Result

Recompute other trigonometric functions:- \(\sec t = \frac{1}{\cos t} = \frac{5}{4}\)- \(\csc t = \frac{1}{\sin t} = -\frac{5}{3}\)- \(\cot t = \frac{1}{\tan t} = -\frac{4}{3}\). Ensure all results correspond to the correct quadrant signs; sine and cosecant are negative, others are positive.

Key Concepts

Tangent Function

The tangent function, denoted as \( \tan \theta \), is a fundamental trigonometric function. It represents the ratio of the opposite side to the adjacent side in a right-angled triangle. Importantly, the sign of the tangent function depends on the quadrant in which the angle \( \theta \) is located. Here are some key points to consider:

• In the first quadrant, \( \tan \theta \) is positive because both sine and cosine are positive.
• In the second quadrant, \( \tan \theta \) is negative since sine is positive but cosine is negative.
• In the third quadrant, both sine and cosine are negative, making \( \tan \theta \) positive.
• In the fourth quadrant, \( \tan \theta \) is negative because sine is negative and cosine is positive.

Since the problem states \( \tan t = -\frac{3}{4} \), it indicates the angle \( t \) is in either the second or fourth quadrant, where tangent is negative. Given that \( \cos t > 0 \), we place the angle in the fourth quadrant.

Cosine Function

The cosine function, often written as \( \cos \theta \), is another primary trigonometric function. It gives the ratio of the adjacent side to the hypotenuse in a right-angled triangle. An essential feature of cosine is its consistent nature across quadrants:

• In the first and fourth quadrants, \( \cos \theta \) is positive.
• Meanwhile, in the second and third quadrants, \( \cos \theta \) is negative.

In the context of the exercise, because \( \cos t > 0 \), we deduced that angle \( t \) must be in the fourth quadrant. Here, the cosine value calculated was \( \cos t = \frac{4}{5} \), in alignment with the positive nature of cosine in this quadrant.Understanding this behavior helps in determining the signs of other trigonometric functions. The positive cosine in the fourth quadrant indicates how other function values should be considered, such as negative sine or tangent.

Sine Function

The sine function, \( \sin \theta \), relates the opposite side to the hypotenuse in a right triangle. The sign of \( \sin \theta \) changes as follows depending on the quadrant:

• In the first quadrant, \( \sin \theta \) is positive.
• Also positive in the second quadrant.
• In the third and fourth quadrants, \( \sin \theta \) is negative.

Applying this to the problem where \( \tan t = -\frac{3}{4} \) and \( \cos t > 0 \), the sine of \( t \) must then be negative since we have already placed the angle in the fourth quadrant.Thus, the calculated \( \sin t = -\frac{3}{5} \) fits perfectly with the characteristics of sine within this quadrant. This understanding of how sine behaves ensures that we interpret the right triangle correctly and ascertain all related trigonometric values for any given angle.