Problem 35 - Precalculus: A Concise Course

Question

Use the given function value(s), and trigonometric identities (including the cofunction identities), to find the indicated trigonometric functions. \(\cot \alpha=5\) (a) \(\tan \alpha\) (b) \(\csc \alpha\) (c) \(\cot \left(90^{\circ}-\alpha\right)\) (d) \(\cos \alpha\)

Step-by-Step Solution

Answer

\(\tan \alpha = 1/5\), \(\csc \alpha = 1/ ((1/5) \cdot \sqrt{1 - (1/5)^2})\), \(\cot \left(90^{\circ}-\alpha\right) = 1/5\), \(\cos \alpha = \sqrt{1 - (1/5)^2}\)

Step 1 Determine the value of \(\tan \alpha\)

\(\tan \alpha\), which is simply the reciprocal of \(\cot \alpha\), can be found directly by taking the reciprocal of 5: \(\tan \alpha = 1/\cot \alpha = 1/5.\)

Step 2 Determine the value of \(\csc \alpha\)

The cosecant function is defined as the reciprocal of the sine function, \(\csc \alpha = 1/\sin \alpha\). Since we know that \(\tan \alpha = \sin \alpha / \cos \alpha\), we can infer that \(\sin \alpha = \tan \alpha \cdot \cos \alpha\). However, the exact value for cosine isn't given. The relationship between sine and cosine is given by the Pythagorean identity in trigonometry: \(\sin^2 \alpha + \cos^2 \alpha = 1\). Thus, \(\cos \alpha = \sqrt{1 - \sin^2 \alpha}\). We can substitute the value of \(\sin \alpha\) from above: \(\cos \alpha = \sqrt{1 - (1/5)^2}\). And the cosecant function can be calculated: \(\csc \alpha = 1/ \sin \alpha = 1/ ( \tan \alpha \cdot \cos \alpha) = 1/ ((1/5) \cdot \sqrt{1 - (1/5)^2})\).

Step 3 Determine the value of \(\cot \left(90^{\circ}-\alpha\right)\)

This is an application of the cotangent co-function identity, which states \(\cot(90^{\circ}-\alpha) = \tan \alpha \). We have already calculated \(\tan \alpha = 1/5\). So, \(\cot \left(90^{\circ}-\alpha\right) = 1/5 \).

Step 4 Determine the value of \(\cos \alpha \)

We have already worked out the value of \(\cos \alpha\) when working out Step 2: \(\cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - (1/5)^2}\). So, we can use this calculated value directly.

Key Concepts

Cofunction Identities

Cofunction identities are a special set of trigonometric identities that relate the trigonometric functions of complementary angles. Complementary angles are two angles that sum up to 90 degrees or \(\pi/2\) radians.
In terms of trigonometric functions, cofunction identities help us understand how the value of one trigonometric function of an angle relates to another function of its complement.
For example:

• \( \sin(90^{\circ} - \alpha) = \cos \alpha \)
• \( \cos(90^{\circ} - \alpha) = \sin \alpha \)
• \( \tan(90^{\circ} - \alpha) = \cot \alpha \)
• \( \cot(90^{\circ} - \alpha) = \tan \alpha \)
• \( \sec(90^{\circ} - \alpha) = \csc \alpha \)
• \( \csc(90^{\circ} - \alpha) = \sec \alpha \)

This exercise uses the cotangent cofunction identity \( \cot(90^{\circ} - \alpha) = \tan \alpha \). Knowing this identity, we can easily determine \( \cot(90^{\circ} - \alpha) \) once we have \( \tan \alpha \), simplifying the problem-solving process.

Reciprocal Identities

Reciprocal identities provide important relationships between pairs of trigonometric functions by deeming one as the reciprocal of the other. This means the inverse of one function gives us the value of the other.
The main reciprocal relationship can be summarized as:

• \( \sin \alpha = 1/\csc \alpha \) and \( \csc \alpha = 1/\sin \alpha \)
• \( \cos \alpha = 1/\sec \alpha \) and \( \sec \alpha = 1/\cos \alpha \)
• \( \tan \alpha = 1/\cot \alpha \) and \( \cot \alpha = 1/\tan \alpha \)

In the exercise, the identity \( \tan \alpha = 1/\cot \alpha \) was used to find \( \tan \alpha \), given \( \cot \alpha = 5 \). By using this identity, we recognized that \( \tan \alpha = 1/5 \). These identities are particularly useful because they allow you to find the value of one trigonometric function given the value of its reciprocal without needing much computation.

Pythagorean Identity

One of the fundamental identities in trigonometry is the Pythagorean identity. It stems from the Pythagorean theorem and describes the relation between the sine and cosine functions:
\[ \sin^2 \alpha + \cos^2 \alpha = 1 \]
This identity is invaluable because it helps you determine missing trigonometric values when provided with just one known function value, as shown in the original exercise.
For example, if you know the value of \( \sin \alpha \) or \( \cos \alpha \), you can easily compute the other using this identity. In the step-by-step solution, we used this identity to calculate \( \cos \alpha = \sqrt{1 - \sin^2 \alpha} \). This process was used when the tangent value \( \tan \alpha = \sin \alpha / \cos \alpha \) was involved and \( \sin \alpha \) needed to be expressed using \( \tan \alpha \) and \( \cos \alpha \). \( \cos \alpha \) then fitted into the equation \( \cos \alpha = \sqrt{1 - (1/5)^2} \).
Using the Pythagorean identity simplifies conversions between different trig functions, making it a cornerstone of trigonometric evaluations.