How to Verify Trigonometric Identities

Verifying trigonometric identities involves proving that two trigonometric expressions are equivalent for all values where both sides are defined. No single method works for all identities, but systematic approaches can help. This guide provides examples with detailed solutions and helpful strategies for proving trigonometric identities.

Strategies for Verifying Trigonometric Identities

When verifying identities, you can:

  1. Start with the more complex side and simplify it to match the other side
  2. Transform both sides into the same expression
  3. Use fundamental trigonometric identities as building blocks
  4. Apply algebraic techniques like factoring, finding common denominators, or multiplying by conjugates

Examples with Step-by-Step Solutions

Example 1: Basic Identity Verification

Verify the identity: \[ \cos x \cdot \tan x = \sin x \]

Solution

Transform the left side using the identity \( \tan x = \dfrac{\sin x}{\cos x } \):

\[ \cos x \cdot \tan x = \cos x \cdot \frac{\sin x}{\cos x} = \sin x \]

The left side simplifies to the right side, verifying the identity.

Example 2: Product Identity

Verify the identity: \[ \cot x \cdot \sec x \cdot \sin x = 1 \]

Solution

Use the identities \( \cot x = \dfrac{\cos x}{\sin x} \) and \( \sec x = \dfrac{1}{\cos x} \):

\[ \cot x \cdot \sec x \cdot \sin x = \frac{\cos x}{\sin x} \cdot \frac{1}{\cos x} \cdot \sin x = 1 \]

The left side simplifies to 1, confirming the identity.

Example 3: Complex Fraction Identity

Verify the identity: \[ \frac{\cot x - \tan x}{\sin x \cdot \cos x} = \csc^2 x - \sec^2 x \]

Solution

Use the identities \( \cot x = \dfrac{\cos x}{\sin x} \) and \( \tan x = \dfrac{\sin x}{\cos x} \):

\[ \frac{\cot x - \tan x}{\sin x \cdot \cos x} = \frac{\frac{\cos x}{\sin x} - \frac{\sin x}{\cos x}}{\sin x \cdot \cos x} \]

Combine terms in the numerator:

\[ = \frac{\frac{\cos^2 x - \sin^2 x}{\sin x \cos x}}{\sin x \cos x} = \frac{\cos^2 x - \sin^2 x}{\sin^2 x \cos^2 x} \]

Separate into two fractions:

\[ = \frac{\cos^2 x}{\sin^2 x \cos^2 x} - \frac{\sin^2 x}{\sin^2 x \cos^2 x} = \frac{1}{\sin^2 x} - \frac{1}{\cos^2 x} \]

Apply reciprocal identities:

\[ = \csc^2 x - \sec^2 x \]

The identity is verified.

Practice Problems

  1. Verify: \( \sin x + \cos x \cdot \cot x = \csc x \)
  2. Verify: \( \dfrac{\csc x}{1 + \csc x} - \dfrac{\csc x}{1 - \csc x} = 2 \sec^2 x \)

Detailed Solutions

Solution to Problem 1

Starting with the left side:

\[ \sin x + \cos x \cdot \cot x = \sin x + \cos x \cdot \frac{\cos x}{\sin x} \] \[ = \sin x + \frac{\cos^2 x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x} \]

Using the Pythagorean identity:

\[ = \frac{1}{\sin x} = \csc x \]

Solution to Problem 2

Combine the fractions on the left side:

\[ \frac{\csc x}{1 + \csc x} - \frac{\csc x}{1 - \csc x} = \frac{\csc x(1 - \csc x) - \csc x(1 + \csc x)}{(1 + \csc x)(1 - \csc x)} \]

Simplify the numerator:

\[ = \frac{\csc x - \csc^2 x - \csc x - \csc^2 x}{1 - \csc^2 x} = \frac{-2\csc^2 x}{1 - \csc^2 x} \]

Use the identity \( \csc^2 x = \frac{1}{\sin^2 x} \):

\[ = \frac{-2/\sin^2 x}{1 - 1/\sin^2 x} = \frac{-2/\sin^2 x}{(\sin^2 x - 1)/\sin^2 x} = \frac{-2}{\sin^2 x - 1} \]

Since \( \sin^2 x - 1 = -\cos^2 x \):

\[ = \frac{-2}{-\cos^2 x} = \frac{2}{\cos^2 x} = 2\sec^2 x \]

Additional Resources