Problem 55 - Algebra and Trigonometry Real Mathematics, Real People

Question

Use the function value(s) and the trigonometric identities to evaluate each trigonometric function. \(\csc \theta=3, \sec \theta=\frac{3 \sqrt{2}}{4}\) (a) \(\sin \theta\) (b) \(\cos \theta\) (c) \(\tan \theta\) (d) \(\sec \left(90^{\circ}-\theta\right)\)

Step-by-Step Solution

Answer

(a) \(\sin \theta = \frac{1}{3}\) (b) \(\cos \theta = \frac{2\sqrt{2}}{3}\) (c) \(\tan \theta = \frac{\sqrt{2}}{6}\) (d) \(\sec \left(90^{\circ}-\theta\right) = 3\)

Step 1: Compute \(\sin \theta\)

Since \(\csc \theta = \frac{1}{\sin \theta}\), \(\sin \theta\) can be calculated as \(\sin \theta = \frac{1}{\csc \theta} = \frac{1}{3}\).

Step 2: Compute \(\cos \theta\)

We have from the question that \(\sec \theta = \frac{1}{\cos \theta}\). So, we can find \(\cos \theta\) as: \(\cos \theta = \frac{1}{\sec \theta} = \frac{4}{3\sqrt{2}} = \frac{2\sqrt{2}}{3}\) (after multiplying and dividing by \(\sqrt{2}\) to rationalize the denominator).

Step 3: Compute \(\tan \theta\)

Since the definition of tangent is \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), we can substitute our results from steps 1 and 2 to find \(\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{1/3}{2\sqrt{2}/3} = \frac{\sqrt{2}}{6}\).

Step 4: Compute \(\sec \left(90^{\circ}-\theta\right)\)

Using the co-function identity \(\sec \left(90^{\circ}-\theta\right) = \csc \theta\), so \(\sec \left(90^{\circ}-\theta\right) = 3\)

Key Concepts

Cosecant Function

The cosecant function is the reciprocal of the sine function. That means it is defined as \( \csc \theta = \frac{1}{\sin \theta} \).
Understanding the cosecant function helps in solving problems where you need to find the sine from a given cosecant value, as seen in our exercise.

• If \( \csc \theta = 3 \), then \( \sin \theta \) can be found by taking the reciprocal, which means \( \sin \theta = \frac{1}{3} \).
• The cosecant function is periodic with a period of 360 degrees or \( 2\pi \) radians.

The cosecant function plays an essential role in trigonometry because it allows transformation between different trigonometric functions using their identities.

Secant Function

Like the cosecant, the secant function is also a reciprocal function. It is the reciprocal of the cosine function, defined as \( \sec \theta = \frac{1}{\cos \theta} \).
This function helps us calculate cosine values easily when the secant is known.

• Given \( \sec \theta = \frac{3\sqrt{2}}{4} \), to find \( \cos \theta \), you take the reciprocal, so \( \cos \theta = \frac{4}{3\sqrt{2}} \).
• After rationalizing the denominator by multiplying and dividing by \( \sqrt{2} \), we get \( \cos \theta = \frac{2\sqrt{2}}{3} \).

The secant function is crucial for its use in co-function identities and its periodicity, mirroring that of the cosine, with the same period of 360 degrees or \( 2\pi \) radians.

Tangent Function

The tangent function, noted as \( \tan \theta \), is the ratio of the sine function to the cosine function. This is defined by the identity \( \tan \theta = \frac{\sin \theta}{\cos \theta} \).
By knowing values for \( \sin \theta \) and \( \cos \theta \), you can calculate \( \tan \theta \).

• In our exercise, since \( \sin \theta = \frac{1}{3} \) and \( \cos \theta = \frac{2\sqrt{2}}{3} \), we substitute these into the tangent formula to get \( \tan \theta = \frac{1/3}{2\sqrt{2}/3} = \frac{\sqrt{2}}{6} \).

The tangent function has a period of 180 degrees or \( \pi \) radians, which is different from cosecant and secant, reflecting its unique role in trigonometry.

Co-function Identities

Co-function identities are equations that relate trigonometric functions of complementary angles. Complementary angles add up to 90 degrees or \( \pi/2 \) radians.
For example, \( \sec(90^\circ - \theta) = \csc \theta \), \( \csc(90^\circ - \theta) = \sec \theta \), and similarly for sine and cosine.

• In our problem, we use \( \sec(90^\circ - \theta) = \csc \theta \) to find that \( \sec(90^\circ - \theta) = 3 \).

These identities are immensely helpful when you need to switch between functions in equations, especially in contexts where angles are complementary.