Evaluating Trigonometric Expressions with Complementary Angles

This problem asks us to find the value of a trigonometric expression involving ratios of cosine and sine of different angles. The expression is: \( \frac{\cos41^\circ}{\sin49^\circ} + \frac{\sin51^\circ}{\cos39^\circ} \).

To solve this, we need to use the concept of complementary angles in trigonometry. Two angles are called complementary if their sum is \(90^\circ\). The key trigonometric identities involving complementary angles are:

• \( \sin(90^\circ - \theta) = \cos \theta \)
• \( \cos(90^\circ - \theta) = \sin \theta \)
• \( \tan(90^\circ - \theta) = \cot \theta \)
• \( \cot(90^\circ - \theta) = \tan \theta \)
• \( \sec(90^\circ - \theta) = \csc \theta \)
• \( \csc(90^\circ - \theta) = \sec \theta \)

Step-by-Step Evaluation of the Expression

Evaluating the First Term: \( \frac{\cos41^\circ}{\sin49^\circ} \)

Let's look at the angles in the first term, \(41^\circ\) and \(49^\circ\). We notice that \(41^\circ + 49^\circ = 90^\circ\). This means \(41^\circ\) and \(49^\circ\) are complementary angles.

We can rewrite \(49^\circ\) as \(90^\circ - 41^\circ\). Using the complementary angle identity \( \sin(90^\circ - \theta) = \cos \theta \), we can write:

\( \sin49^\circ = \sin(90^\circ - 41^\circ) = \cos41^\circ \)

Now, substitute this back into the first term:

\( \frac{\cos41^\circ}{\sin49^\circ} = \frac{\cos41^\circ}{\cos41^\circ} \)

Assuming \( \cos41^\circ \neq 0 \) (which is true), we can cancel out \( \cos41^\circ \) from the numerator and denominator:

\( \frac{\cos41^\circ}{\cos41^\circ} = 1 \)

So, the value of the first term is 1.

Evaluating the Second Term: \( \frac{\sin51^\circ}{\cos39^\circ} \)

Now consider the angles in the second term, \(51^\circ\) and \(39^\circ\). We notice that \(51^\circ + 39^\circ = 90^\circ\). This means \(51^\circ\) and \(39^\circ\) are complementary angles.

We can rewrite \(39^\circ\) as \(90^\circ - 51^\circ\). Using the complementary angle identity \( \cos(90^\circ - \theta) = \sin \theta \), we can write:

\( \cos39^\circ = \cos(90^\circ - 51^\circ) = \sin51^\circ \)

Now, substitute this back into the second term:

\( \frac{\sin51^\circ}{\cos39^\circ} = \frac{\sin51^\circ}{\sin51^\circ} \)

Assuming \( \sin51^\circ \neq 0 \) (which is true), we can cancel out \( \sin51^\circ \) from the numerator and denominator:

\( \frac{\sin51^\circ}{\sin51^\circ} = 1 \)

So, the value of the second term is also 1.

Combining the Results

The original expression is the sum of the two terms we just evaluated:

\( \frac{\cos41^\circ}{\sin49^\circ} + \frac{\sin51^\circ}{\cos39^\circ} = 1 + 1 = 2 \)

Therefore, the value of the given trigonometric expression is 2.

Revision Table: Key Trigonometric Identities

Additional Information: Understanding Complementary Angles in Trigonometry

The relationship between trigonometric ratios of complementary angles is fundamental in trigonometry. It arises directly from the properties of right-angled triangles.

Consider a right-angled triangle ABC, right-angled at B. Let \( \angle BAC = \theta \). Then the other acute angle, \( \angle BCA \), is \(90^\circ - \theta\).

In this triangle:

• \( \sin \theta = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{BC}{AC} \)
• \( \cos \theta = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{AB}{AC} \)
• \( \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{BC}{AB} \)

Now, let's look at the angle \(90^\circ - \theta\):

• \( \sin(90^\circ - \theta) = \frac{\text{Opposite side to }(90^\circ - \theta)}{\text{Hypotenuse}} = \frac{AB}{AC} \)
• \( \cos(90^\circ - \theta) = \frac{\text{Adjacent side to }(90^\circ - \theta)}{\text{Hypotenuse}} = \frac{BC}{AC} \)
• \( \tan(90^\circ - \theta) = \frac{\text{Opposite side to }(90^\circ - \theta)}{\text{Adjacent side to }(90^\circ - \theta)} = \frac{AB}{BC} \)

Comparing these ratios:

• We see that \( \sin(90^\circ - \theta) = \frac{AB}{AC} \) and \( \cos \theta = \frac{AB}{AC} \). Therefore, \( \sin(90^\circ - \theta) = \cos \theta \).
• We see that \( \cos(90^\circ - \theta) = \frac{BC}{AC} \) and \( \sin \theta = \frac{BC}{AC} \). Therefore, \( \cos(90^\circ - \theta) = \sin \theta \).
• We see that \( \tan(90^\circ - \theta) = \frac{AB}{BC} \) and \( \cot \theta = \frac{1}{\tan \theta} = \frac{1}{(BC/AB)} = \frac{AB}{BC} \). Therefore, \( \tan(90^\circ - \theta) = \cot \theta \).

These relationships are very useful for simplifying trigonometric expressions and solving equations involving complementary angles, as demonstrated in the solution to this problem.