Problem 57 - Algebra and Trigonometry with Analytic Geometry

Question

Verify the identity by transforming the lefthand side into the right-hand side. $$\cos ^{2} \theta\left(\sec ^{2} \theta-1\right)=\sin ^{2} \theta$$

Step-by-Step Solution

Answer

Identity verified: \( \cos^2 \theta (\sec^2 \theta - 1) = \sin^2 \theta \).

Step 1: Write Down the Given Identity

The identity to verify is \( \cos^2 \theta (\sec^2 \theta - 1) = \sin^2 \theta \). We need to show that the left-hand side can be transformed into the right-hand side.

Step 2: Simplify the Left-Hand Side Expression

The left-hand side is \( \cos^2 \theta (\sec^2 \theta - 1) \). Start by using the identity \( \sec \theta = \frac{1}{\cos \theta} \), which implies \( \sec^2 \theta = \frac{1}{\cos^2 \theta} \). Substituting, we have:\[ \cos^2 \theta \left( \frac{1}{\cos^2 \theta} - 1 \right) \]

Step 3: Simplify Inside the Parentheses

Now simplify the expression inside the parentheses:\[ \frac{1}{\cos^2 \theta} - 1 = \frac{1 - \cos^2 \theta}{\cos^2 \theta} \]Substitute back into the main expression: \[ \cos^2 \theta \cdot \frac{1 - \cos^2 \theta}{\cos^2 \theta} \]

Step 4: Cancel Out \( \cos^2 \theta \)

Since \( \cos^2 \theta \) is in both the numerator and the denominator, they cancel each other:\[ 1 - \cos^2 \theta \]

Step 5: Use Pythagorean Identity

Now apply the Pythagorean identity \( \sin^2 \theta = 1 - \cos^2 \theta \). Thus,\[ 1 - \cos^2 \theta = \sin^2 \theta \]

Step 6: Conclusion

Since we have shown that the left-hand side simplifies to \( \sin^2 \theta \), the given identity is verified. Thus, \( \cos^2 \theta (\sec^2 \theta - 1) = \sin^2 \theta \).

Key Concepts

Pythagorean Identity

One of the fundamental identities in trigonometry is the Pythagorean Identity, which is crucial for simplifying expressions involving trigonometric functions. It states that for any angle \( \theta \): \[ \sin^2 \theta + \cos^2 \theta = 1 \] This identity indicates the intrinsic relationship between sine and cosine functions. This is a direct result of the Pythagorean theorem, applied in the context of the unit circle where the radius is always 1.

• It can be visualized by considering a right triangle drawn in the unit circle on a Cartesian plane.
• The hypotenuse, radius of the circle, is \(1\), leading to the equation for the sum of the squares of sine and cosine being equal to one.

The Pythagorean identity can be rearranged to derive other identities, like \( \sin^2 \theta = 1 - \cos^2 \theta \) and \( \cos^2 \theta = 1 - \sin^2 \theta \). This is exactly what we used in this exercise to demonstrate our initial equality.

Sine Function

The sine function is one of the primary building blocks of trigonometry. For a given angle \( \theta \), the sine function relates to the y-coordinate of a point on the unit circle. Mathematically, the sine of angle \( \theta \) is the ratio of the length of the opposite side to the length of the hypotenuse in a right triangle: \[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} \] The function is periodic with a period of \(2\pi\), meaning it repeats its values in a regular pattern every \(2\pi\) radians. Sine function values range between -1 and 1.

• It is an odd function, which implies \( \sin(-\theta) = -\sin(\theta) \).
• It reaches its maximum value of 1 when \( \theta = \frac{\pi}{2} + 2k\pi \), where \(k\) is an integer.

The sine function is crucial for solving the given exercise because it helps in expressing the result of the identity transformation, showing that \( \sin^2 \theta \) can be the final expression derived from the original complex identity.

Cosine Function

Like the sine function, the cosine function is a foundational element in trigonometry. Cosine of an angle \( \theta \) is the x-coordinate of a point on the unit circle, or the ratio of the adjacent side to the hypotenuse in a right triangle: \[ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} \] This function also has a period of \(2\pi\) and its values range from -1 to 1.

• It is an even function, meaning \( \cos(-\theta) = \cos(\theta) \).
• It achieves its maximum value of 1 at \( \theta = 2k\pi \), where \(k\) is an integer.

In this exercise, understanding the properties of the cosine function helped in manipulating \( \cos^2 \theta \). It's with the property of the cosine function that we used identities like \( \sec^2 \theta = \frac{1}{\cos^2 \theta} \) to simplify the expressions.

Secant Function

The secant function is less commonly discussed compared to sine and cosine, but it plays an important role in trigonometry. It is the reciprocal of the cosine function: \[ \sec \theta = \frac{1}{\cos \theta} \] This relationship makes secant undefined for angles where cosine equals zero (i.e., \( \frac{\pi}{2} + k\pi \) where \(k\) is an integer).

• The secant function is periodic with a period of \(2\pi\).
• Because it is the reciprocal of an even function, secant is also even: \( \sec(-\theta) = \sec(\theta) \).

In the exercise, we utilized the identity \( \sec^2 \theta = \frac{1}{\cos^2 \theta} \) to rework the original expression. This crucial step allowed us to apply further simplifications, leading us to appreciate how secant interplays with other trigonometric functions to verify the given identity successfully.