Problem 10 - Algebra and Trigonometry

Question

\(3-10\) Find \(\sin 2 x, \cos 2 x,\) and \(\tan 2 x\) from the given information. $$ \cot x=\frac{2}{3}, \quad \sin x>0 $$

Step-by-Step Solution

Answer

\( \sin 2x = \frac{12}{13} \), \( \cos 2x = -\frac{5}{13} \), \( \tan 2x = -\frac{12}{5} \).

Step 1: Recall identities for double angles

The double angle formulas are crucial in this problem. We use:- \( \sin(2x) = 2\sin(x)\cos(x) \)- \( \cos(2x) = \cos^2(x) - \sin^2(x) \)- \( \tan(2x) = \frac{2\tan(x)}{1 - \tan^2(x)} \).

Step 2: Convert \( \cot x \) to \( \tan x \)

Given \( \cot x = \frac{2}{3} \), use the identity \( \tan x = \frac{1}{\cot x} \). Therefore, \( \tan x = \frac{3}{2} \).

Step 3: Find \( \sin x \) and \( \cos x \) from \( \tan x \)

Since \( \tan x = \frac{\sin x}{\cos x} = \frac{3}{2} \), assume \( \sin x = 3k \) and \( \cos x = 2k \) where \( k \) is a common multiple. Using \( \sin^2 x + \cos^2 x = 1 \), substitute to get:\((3k)^2 + (2k)^2 = 1\)\(9k^2 + 4k^2 = 1\)\(13k^2 = 1\)\(k = \frac{1}{\sqrt{13}} \).

Step 4: Calculate \( \sin x \) and \( \cos x \)

Using \( k = \frac{1}{\sqrt{13}} \), find:\( \sin x = 3\cdot \frac{1}{\sqrt{13}} = \frac{3}{\sqrt{13}} \)\( \cos x = 2\cdot \frac{1}{\sqrt{13}} = \frac{2}{\sqrt{13}} \)

Step 5: Compute \( \sin 2x \)

Using \( \sin(2x) = 2\sin x \cos x \):\[ \sin(2x) = 2 \left( \frac{3}{\sqrt{13}} \right) \left( \frac{2}{\sqrt{13}} \right) = 2 \cdot \frac{6}{13} = \frac{12}{13} \]

Step 6: Compute \( \cos 2x \)

Using \( \cos(2x) = \cos^2 x - \sin^2 x \):\[ \cos(2x) = \left( \frac{2}{\sqrt{13}} \right)^2 - \left( \frac{3}{\sqrt{13}} \right)^2 = \frac{4}{13} - \frac{9}{13} = -\frac{5}{13} \]

Step 7: Compute \( \tan 2x \)

Using \( \tan(2x) = \frac{2 \tan x}{1 - \tan^2 x} \):\( \tan^2 x = \left( \frac{3}{2} \right)^2 = \frac{9}{4} \).\[ \tan(2x) = \frac{2 \cdot \frac{3}{2}}{1 - \frac{9}{4}} = \frac{3}{\frac{-5}{4}} = -\frac{12}{5} \]

Key Concepts

Double Angle Identities

The double angle identities are important tools in trigonometry that allow us to express trigonometric functions of double angles in terms of single angles. They serve many practical purposes, such as simplifying expressions or solving trigonometric equations.
\[ \text{Here are the key double angle identities:} \]

• For sine: \( \sin(2x) = 2\sin(x)\cos(x) \)
• For cosine: \( \cos(2x) = \cos^2(x) - \sin^2(x) \)
• For tangent: \( \tan(2x) = \frac{2\tan(x)}{1 - \tan^2(x)} \)

These identities allow us to find \( \sin(2x) \), \( \cos(2x) \), and \( \tan(2x) \) when \( \sin(x) \), \( \cos(x) \), or \( \tan(x) \) are known. They are particularly useful when you have angles like \( 2x \) that are double some known value \( x \). Understanding these identities helps in problem-solving by reducing complex trigonometric equations to simpler forms. The identities work based on fundamental trigonometric definitions and can be derived from basic trigonometry principles.

Cotangent to Tangent Conversion

Converting between cotangent and tangent is a basic yet crucial skill in trigonometry. This conversion is especially relevant when you need to find values of one function given the other.
This transformation is grounded in their mathematical relationship:

• The cotangent function is the reciprocal of the tangent function: \( \cot(x) = \frac{1}{\tan(x)} \)
• Thus, to convert \( \cot(x) = \frac{2}{3} \) to \( \tan(x) \), you simply take the reciprocal: \( \tan(x) = \frac{3}{2} \)

When facing problems like these, remember that reciprocal functions are straightforward; switching from cotangent to tangent involves inverting the fraction. This practice is especially shared in trigonometric identities and equations where multiple function conversions occur.
Understanding this conversion supports solving more intricate trigonometric expressions and enhances flexibility when working with these functions. It is a key feature of many trigonometric exercises, smoothing the path to finding solutions.

Pythagorean Identity

The Pythagorean identity is one of the cornerstones of trigonometry, closely related to the Pythagorean theorem from geometry. This identity plays a vital role in connecting sine and cosine functions:
\[ \sin^2(x) + \cos^2(x) = 1 \]
This equation holds true for any angle \( x \), allowing you to find one trigonometric function based on the other.
In practical use, suppose you find through another identity that \( \tan(x) = \frac{3}{2} \). Assuming \( \sin(x) = 3k \) and \( \cos(x) = 2k \), you can apply the Pythagorean identity:

• Substitute and solve: \( (3k)^2 + (2k)^2 = 1 \)
• Simplify to find \( k = \frac{1}{\sqrt{13}} \)
• Then calculate: \( \sin(x) = \frac{3}{\sqrt{13}} \) and \( \cos(x) = \frac{2}{\sqrt{13}} \)

This identity is central in figuring out the values of trigonometric functions that conform to the unit circle model. It allows you to deduce unknown values and supports various trigonometric transformations, providing a reliable method to check steps in trigonometric problem-solving.