Problem 12 - Contemporary Precalculus

Question

In Exercises \(7-16,\) evaluate all six trigonometric finctions at \(t\) where the given point lies on the terminal side of an angle of \(t\) radians in standard position. $$(4 / 5,-3 / 5)$$

Step-by-Step Solution

Answer

The six trigonometric functions evaluated at angle \(t\) are: - sine: \(\sin t = -\frac{3}{5}\). - cosine: \(\cos t = \frac{4}{5}\). - tangent: \(\tan t = -\frac{3}{4}\). - cosecant: \(\csc t = -\frac{5}{3}\). - secant: \(\sec t = \frac{5}{4}\). - cotangent: \(\cot t = -\frac{4}{3}\).

Step 1: Identify the given values

The given point is \((4/5, -3/5)\). So, we have the values of \(x\) and \(y\), i.e., \(x = 4/5\) and \(y = -3/5\).

Step 2: Calculate the length of the hypotenuse (r)

We will use the Pythagorean theorem to find the length of the hypotenuse, \(r\): \(r^2 = x^2 + y^2\). Plug in the values of \(x\) and \(y\): \(r^2 = (4/5)^2 + (-3/5)^2 = 16/25 + 9/25 = 25/25\). Taking the square root of both sides, we get \(r = 1\).

Step 3: Find sine, cosine, and tangent of the angle (t)

For the angle (t), we have the following: - sine: \(\sin t = \frac{y}{r} = \frac{-3/5}{1} = -\frac{3}{5}\). - cosine: \(\cos t = \frac{x}{r} = \frac{4/5}{1} = \frac{4}{5}\). - tangent: \(\tan t = \frac{y}{x} = \frac{-3/5}{4/5} = -\frac{3}{4}\).

Step 4: Find the reciprocal trigonometric functions

- cosecant: \(\csc t = \frac{1}{\sin(t)} = \frac{1}{-3/5} = -\frac{5}{3}\). - secant: \(\sec t = \frac{1}{\cos(t)} = \frac{1}{4/5} = \frac{5}{4}\). - cotangent: \(\cot t = \frac{1}{\tan(t)} = \frac{1}{-3/4} = -\frac{4}{3}\). The six trigonometric functions evaluated at angle \(t\) are: - sine: \(\sin t = -\frac{3}{5}\). - cosine: \(\cos t = \frac{4}{5}\). - tangent: \(\tan t = -\frac{3}{4}\). - cosecant: \(\csc t = -\frac{5}{3}\). - secant: \(\sec t = \frac{5}{4}\). - cotangent: \(\cot t = -\frac{4}{3}\).

Key Concepts

Trigonometry

Trigonometry is a fascinating area of mathematics that intertwines geometry and algebra to solve problems involving angles and distances. It investigates the properties of triangles, particularly the relationships between the angles and the lengths of sides of right triangles. In the context of our exercise, trigonometry provides the tools to evaluate the trigonometric functions, which are essential in defining the relationships between the angle of a triangle and the ratios of its sides.

For a point \( (x, y) \) lying on the terminal side of an angle \( t \) in standard position, we can define trigonometric functions such as sine (sin), cosine (cos), and tangent (tan) based on the values of \( x \) and \( y \) and the hypotenuse \( r \). Specifically, \( \sin(t) \) is the ratio of the opposite side to the hypotenuse, \( \cos(t) \) is the ratio of the adjacent side to the hypotenuse, and \( \tan(t) \) is the ratio of the opposite side to the adjacent side. These basic functions allow mathematicians and scientists to calculate angles and distances in a variety of disciplines, ranging from physics and engineering to even fields like astronomy and geology.

Pythagorean theorem

The Pythagorean theorem is undoubtedly one of the most well-known results in mathematics, named after the ancient Greek mathematician Pythagoras. It states that in a right-angled triangle, the square of the length of the hypotenuse \( (r) \) is equal to the sum of the squares of the lengths of the other two sides. The theorem can be written as \( a^2 + b^2 = r^2 \) where \( a \) and \( b \) are the lengths of the legs of the triangle, and \( r \) is the length of the hypotenuse.

In our exercise, the Pythagorean theorem played a key role in determining the hypotenuse of the triangle formed by the point \( (4/5, -3/5) \) and the origin. By applying the theorem, \( r^2 = (4/5)^2 + (-3/5)^2 \) simplifies to \( r = 1 \) after taking the square root of both sides. This insight is crucial for evaluating trigonometric functions accurately, as \( r \) is often the denominator in their ratios.

Reciprocal trigonometric functions

In addition to the primary trigonometric functions, there are also three reciprocal trigonometric functions, which are derived from them. These are the cosecant \( (\csc) \) which is the reciprocal of the sine, the secant \( (\sec) \) which is the reciprocal of the cosine, and the cotangent \( (\cot) \) which is the reciprocal of the tangent. Such functions often provide convenience and simplifications in trigonometric expressions and equations.

To clarify, if we have \( \sin(t) = -\frac{3}{5} \) and \( \cos(t) = \frac{4}{5} \) as with our point \( (4/5, -3/5) \) from the exercise, their reciprocals would be calculated as \( \csc(t) = \frac{1}{\sin(t)} = -\frac{5}{3} \) and \( \sec(t) = \frac{1}{\cos(t)} = \frac{5}{4} \) respectively. Similarly, if \( \tan(t) = -\frac{3}{4} \) then \( \cot(t) = \frac{1}{\tan(t)} = -\frac{4}{3} \). Understanding these relationships can vastly improve a student’s ability to work with trigonometric identities and solve trigonometric equations, which are applications found throughout various branches of mathematics and physics.