Problem 103 - Precalculus

Question

For the following exercises, find the exact values of a \() \sin (2 x),\) b) \(\cos (2 x),\) and \(c ) \tan (2 x)\) without solving for \(x\) If \(\sin x=\frac{1}{8}, \quad\) and \(x\) is in quadrant I.

Step-by-Step Solution

Answer

\(\sin(2x) = \frac{\sqrt{63}}{32}, \cos(2x) = \frac{31}{32}, \tan(2x) = \frac{1}{\sqrt{63}}\).

Step 1: Recall Double Angle Formulas

To find the exact values of \( \sin(2x) \), \( \cos(2x) \), and \( \tan(2x) \), we need to use the double angle formulas: \( \sin(2x) = 2 \sin x \cos x \), \( \cos(2x) = \cos^2 x - \sin^2 x \), and \( \tan(2x) = \frac{2\tan x}{1-\tan^2 x} \). Since \( x \) is in the first quadrant, both \( \sin x \) and \( \cos x \) are positive.

Step 2: Find \(\cos x\)

We know that \( \sin x = \frac{1}{8} \). Since \( x \) is in the first quadrant, use the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \) to find \( \cos x \). Calculate \( \cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - \left(\frac{1}{8}\right)^2} = \sqrt{\frac{63}{64}} = \frac{\sqrt{63}}{8} \).

Step 3: Calculate \(\sin(2x)\)

Using the double angle formula for sine, \( \sin(2x) = 2 \sin x \cos x \), substitute \( \sin x = \frac{1}{8} \) and \( \cos x = \frac{\sqrt{63}}{8} \). Compute \( \sin(2x) = 2 \times \frac{1}{8} \times \frac{\sqrt{63}}{8} = \frac{\sqrt{63}}{32} \).

Step 4: Calculate \(\cos(2x)\)

Using the double angle formula for cosine, \( \cos(2x) = \cos^2 x - \sin^2 x \), substitute the values \( \cos x = \frac{\sqrt{63}}{8} \) and \( \sin x = \frac{1}{8} \). Compute \( \cos^2 x = \frac{63}{64} \) and \( \sin^2 x = \frac{1}{64} \). Thus, \( \cos(2x) = \frac{63}{64} - \frac{1}{64} = \frac{62}{64} = \frac{31}{32} \).

Step 5: Calculate \(\tan(2x)\)

Using the double angle formula for tangent, \( \tan(2x) = \frac{2\tan x}{1 - \tan^2 x} \), first find \( \tan x = \frac{\sin x}{\cos x} = \frac{\frac{1}{8}}{\frac{\sqrt{63}}{8}} = \frac{1}{\sqrt{63}} \). Thus, \( \tan^2 x = \frac{1}{63} \). Substitute these into the formula to get \( \tan(2x) = \frac{2 \times \frac{1}{\sqrt{63}}}{1 - \frac{1}{63}} = \frac{\frac{2}{\sqrt{63}}}{\frac{62}{63}} = \frac{2 \times 63}{\sqrt{63} \times 62} = \frac{126}{62\sqrt{63}} = \frac{63}{31\sqrt{63}} = \frac{1}{\sqrt{63}} \).

Key Concepts

Double Angle Formulas

The double angle formulas are fundamental tools in trigonometry that allow us to express trigonometric functions of double angles in terms of single angles. These formulas are particularly useful when you want to find values for angles like \( 2x \) using known values of trigonometric functions at \( x \). The main double angle formulas are:

• \( \sin(2x) = 2 \sin x \cos x \)
• \( \cos(2x) = \cos^2 x - \sin^2 x \)
• \( \tan(2x) = \frac{2 \tan x}{1 - \tan^2 x} \)

These identities simplify our calculations significantly. For example, when given \( \sin x \) and \( \cos x \) for a specific angle \( x \), you can directly find \( \sin(2x) \), \( \cos(2x) \), and \( \tan(2x) \) without first determining the value of \( x \). The formulas depend on knowing the sine, cosine, or tangent of the original angle \( x \), which can often be determined using other trigonometric identities.

Pythagorean Identity

The Pythagorean identity is one of the most essential trigonometric identities. It connects the squares of sine and cosine of an angle to 1:

• \( \sin^2 x + \cos^2 x = 1 \)

This identity stems from the Pythagorean Theorem applied to the unit circle, where any point \((\cos x, \sin x)\) lies on a circle of radius 1. It's particularly useful for calculating one trigonometric function given another at the same angle.
In our exercise, since \( \sin x = \frac{1}{8} \) and \( x \) is in the first quadrant, we are assured the cosine is positive. To find \( \cos x \), rearrange the identity to \( \cos x = \sqrt{1 - \sin^2 x} \), leading us to the value \( \frac{\sqrt{63}}{8} \). This is crucial for applying double angle formulas.

Quadrant

In trigonometry, the quadrant of an angle is crucial when considering the signs of sine, cosine, and tangent. The unit circle is divided into four quadrants:

• Quadrant I: Both \( \sin x \) and \( \cos x \) are positive.
• Quadrant II: \( \sin x \) is positive, \( \cos x \) is negative.
• Quadrant III: Both \( \sin x \) and \( \cos x \) are negative.
• Quadrant IV: \( \sin x \) is negative, \( \cos x \) is positive.

For angles in quadrant I, like our exercise with \( \sin x = \frac{1}{8} \), both trigonometric components (sine and cosine) retain their positive values. This ensures simple application of identities without negation. Knowing the quadrant framework helps you understand the behavior of trigonometric functions within different sections of the circle, helping you select and use the correct signs for your calculations.

Exact Values

Exact values in trigonometry refer to precise expressions of trigonometric functions, as opposed to decimal approximations. They are particularly important because they provide more accurate calculations and interpretations.
Using identities and formulas, we often express exact values in terms of radicals or fractions, like how the given \( \sin x = \frac{1}{8} \) helped determine \( \cos x = \frac{\sqrt{63}}{8} \) using the Pythagorean identity.
By substituting these into the double angle formulas, you can find exact values of \( \sin(2x) \), \( \cos(2x) \), and \( \tan(2x) \) as fractions or expressions involving roots, retaining their high precision. Mastering how to compute and understand exact values is valuable, especially in settings where approximation errors might impact results significantly.