Problem 7 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-50: Verify the identity. $$ \frac{1+\cos 3 t}{\sin 3 t}+\frac{\sin 3 t}{1+\cos 3 t}=2 \csc 3 t $$

Step-by-Step Solution

Answer

The given identity is verified as correct.

Step 1: Express Both Sides in Terms of Sine and Cosine

First, observe the given expression on the left side: \[ \frac{1+\cos 3t}{\sin 3t} + \frac{\sin 3t}{1+\cos 3t} \]The right side of the identity is expressed with the cosecant function, which is \(\csc 3t = \frac{1}{\sin 3t}\). Our goal is to simplify the left side to match the right side expression.

Step 2: Find a Common Denominator

To simplify the left side, we'll find a common denominator for the two fractions:\[ \text{Common Denominator} = \sin 3t (1 + \cos 3t) \]Write: \[ \frac{1+\cos 3t}{\sin 3t} = \frac{(1+\cos 3t)^2}{\sin 3t(1+\cos 3t)} \]\[ \frac{\sin 3t}{1+\cos 3t} = \frac{(\sin 3t)^2}{\sin 3t(1+\cos 3t)} \]

Step 3: Combine the Fractions

Combine the fractions over the common denominator:\[ \frac{(1+\cos 3t)^2 + (\sin 3t)^2}{\sin 3t (1+\cos 3t)} \]Apply the Pythagorean identity: \((\sin 3t)^2 + (\cos 3t)^2 = 1\).

Step 4: Simplify the Numerator Using Trigonometric Identity

Substitute the identity in the numerator:\[ (1 + 2\cos 3t + \cos^2 3t) + (\sin^2 3t) \]Simplify using \(\cos^2 3t + \sin^2 3t = 1\):\[ 1 + 2\cos 3t + 1 = 2 + 2\cos 3t \]

Step 5: Factor and Simplify Further

Factor the numerator:\[ (1+\cos 3t)^2 = 1 + 2\cos 3t + \cos^2 3t \]Therefore,\[ \frac{2(1+\cos 3t)}{\sin 3t (1+\cos 3t)} \]Cancel \(1+\cos 3t\) from numerator and denominator:\[ \frac{2}{\sin 3t} = 2\csc 3t \]

Step 6: Verify the Identity

We've simplified the left-hand side of the expression to match the right-hand side \(2 \csc 3t\). This verifies the given trigonometric identity is indeed correct.

Key Concepts

Sine and Cosine Functions

The sine and cosine functions are fundamental in trigonometry. They describe relationships in right-angled triangles and the unit circle. You can visualize them as the y and x coordinates of a point rotating around a circle. These functions repeat their values in a periodic fashion every 360 degrees, or \(2\pi\) radians.

• Sine Function (\(\sin\)): This function gives the vertical component. It's defined as the ratio of the opposite side to the hypotenuse in a right triangle. Its values range from -1 to 1.
• Cosine Function (\(\cos\)): This function provides the horizontal component. It's the ratio of the adjacent side to the hypotenuse. Like sine, its values are also between -1 and 1.

Sine and cosine are often used together in various trigonometric identities. They help us transform and simplify expressions, making them a crucial aspect of trigonometry.

Cosecant Function

The cosecant function, denoted as \(\csc\), is the reciprocal of the sine function. While sine refers to the ratio of the opposite side to the hypotenuse, cosecant relates to the hypotenuse over the opposite side.

• Mathematically, \(\csc \theta = \frac{1}{\sin \theta}\).
• The cosecant function is undefined whenever \(\sin \theta = 0\), as division by zero is not possible.

In trigonometric identities, cosecant often pops up when we need to express certain terms in simpler forms, as seen in expressions where fractions involving sine are transformed. Understanding and using cosecant allows for solutions that maintain consistency across problems.

Pythagorean Identity

The Pythagorean identity is one of the most important relationships in trigonometry. It states that the square of the sine function plus the square of the cosine function equals one, i.e., \(\sin^2 \theta + \cos^2 \theta = 1\).

• This identity arises from the Pythagorean theorem, applied within the context of a unit circle.
• It is foundational for simplifying and proving other trigonometric identities.
• When encountering expressions like in the original exercise, using the Pythagorean identity can help reduce complex expressions into simpler terms.

The identity not only serves in verifying trigonometric problems but also in discovering new relationships and simplifying mathematical expressions in calculus and beyond.