Problem 13 - A Graphical Approach to Precalculus with Limits

Question

Use identities to find (a) \(\sin 2 \theta\) and (b) \(\cos 2 \theta\) $$\tan \theta=2 \text { and } \cos \theta>0$$

Step-by-Step Solution

Answer

(a) \(\sin 2\theta = \frac{4}{5}\) and (b) \(\cos 2\theta = -\frac{3}{5}\).

Step 1: Understand the Trigonometric Identity

Recall the double angle identities: \( \sin 2\theta = 2 \sin \theta \cos \theta \) and \( \cos 2\theta = \cos^2 \theta - \sin^2 \theta \). We need to find \( \sin \theta \) and \( \cos \theta \) first using the given \( \tan \theta \).

Step 2: Use the Tangent Identity

Since \( \tan \theta = 2 \), we know \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). Thus, \( \sin \theta = 2 \cos \theta \).

Step 3: Apply Pythagorean Identity

From the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \), substitute \( \sin \theta = 2 \cos \theta \) into the equation: \((2 \cos \theta)^2 + \cos^2 \theta = 1\).

Step 4: Solve for \( \cos \theta \)

Simplify to obtain: \(4 \cos^2 \theta + \cos^2 \theta = 1\), which becomes \(5 \cos^2 \theta = 1\). Thus, \( \cos \theta = \sqrt{\frac{1}{5}} \) (since \( \cos \theta > 0 \)).

Step 5: Solve for \( \sin \theta \)

Using \( \sin \theta = 2 \cos \theta \), substitute \( \cos \theta = \sqrt{\frac{1}{5}} \) to find \( \sin \theta = 2\sqrt{\frac{1}{5}} = \sqrt{\frac{4}{5}} \).

Step 6: Calculate \( \sin 2\theta \)

Use the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \). Substitute the values: \( \sin 2\theta = 2 \times \sqrt{\frac{4}{5}} \times \sqrt{\frac{1}{5}} = \frac{4}{5} \).

Step 7: Calculate \( \cos 2\theta \)

Use the identity \( \cos 2\theta = \cos^2 \theta - \sin^2 \theta \). Substitute the values: \( \cos 2\theta = \left(\frac{1}{5}\right) - \left(\frac{4}{5}\right) = -\frac{3}{5} \).

Key Concepts

Double Angle Formulas

The Double Angle Formulas are fundamental in trigonometry and help us express trigonometric functions of double angles in terms of single angles. They are especially useful when simplifying expressions or solving trigonometric equations. Here are the primary double angle formulas:

• For Sine: \( \sin 2\theta = 2 \sin \theta \cos \theta \)
• For Cosine: \( \cos 2\theta = \cos^2 \theta - \sin^2 \theta \)

These formulas allow you to calculate \( \sin \) and \( \cos \) of an angle that is twice as large without directly measuring the angle itself. In our exercise, we use these formulas to find \( \sin 2\theta \) and \( \cos 2\theta \), given the information about \( \tan \theta \) and the sign of \( \cos \theta \). To solve these, we first determined \( \sin \theta \) and \( \cos \theta \) using given trigonometric identities.

Pythagorean Identity

The Pythagorean Identity is a crucial relation in trigonometry that links the squares of the sine and cosine functions of an angle. It is expressed as:

• \( \sin^2 \theta + \cos^2 \theta = 1 \)

This identity is derived from the Pythagorean Theorem and applies to right triangles. It states that the square of the hypotenuse (equal to 1 in the unit circle) is the sum of the squares of the other two sides, represented by \( \sin \theta \) and \( \cos \theta \).
In the exercise, this identity helps us solve for \( \cos \theta \) after expressing \( \sin \theta \) in terms of \( \cos \theta \) due to the relationship \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). By substituting these into the Pythagorean identity, we reach an equation that allows the calculation of both \( \sin \theta \) and \( \cos \theta \).

Tangent Function

The Tangent Function is another primary trigonometric function, defined as the ratio of sine to cosine:

• \( \tan \theta = \frac{\sin \theta}{\cos \theta} \)

This function is periodic with a period of \( \pi \) and is particularly useful in expressing angles where the opposite side over the adjacent side is known in a right angle triangle.
The given problem begins by using that \( \tan \theta = 2 \). This indicates a specific relationship between \( \sin \theta \) and \( \cos \theta \), which allows us to express one in terms of the other. This is essential for applying the Pythagorean identity in subsequent steps to find both \( \sin \theta \) and \( \cos \theta \).

Sine Function

The Sine Function, one of the basic trigonometric functions, measures the y-coordinate of a point on the unit circle corresponding to an angle \( \theta \). It is defined as:

• \( \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} \)

For our purposes in this exercise, after determining \( \cos \theta \) using the Pythagorean Identity, we used the relationship from the Tangent Function \( \sin \theta = 2 \cos \theta \) to determine \( \sin \theta \).
This simple manipulative step is crucial as it sets the stage for the use of double angle formulas, where \( \sin \theta \) pairs with \( \cos \theta \) to find \( \sin 2\theta \).

Cosine Function

The Cosine Function, along with sine and tangent, makes up the primary trigonometric functions and is essential for understanding circular and oscillatory motion. It is defined as:

• \( \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} \)

In the context of our problem, we determined \( \cos \theta \) by using the Pythagorean Identity after establishing \( \sin \theta = 2 \cos \theta \). With the condition \( \cos \theta > 0 \), we must choose the positive root, resulting in \( \cos \theta = \sqrt{\frac{1}{5}} \). This precise value is then utilized alongside \( \sin \theta \) to apply the double angle formulas, eventually leading to the evaluation of \( \cos 2\theta \).